Can we show that time is orthogonal to space? It's easy to show that the time we measure is "in a different direction" from the space directions we measure. However, it's not immediately obvious to me that these directions are orthogonal.
How do we prove that any directions are orthogonal? What I came up with is the following. [For short distances/does this cause trouble?], we can use the pythagorean theorem. If I have two directions in space, I can simply travel 1 meter in each of the two directions, and then measure the distance between the two endpoints. The result is sqrt(2) meters if and only if your two directions were perpendicular.

Easy!
My question is: can this method be used to prove that our concept of time is orthogonal to our concept of space?
I have an idea of where to start. We would need to move in space by 1 meter, while moving in time by 1 meter (converting time into meters using c). Then we would need to measure whether the distance traveled was sqrt(2) meters. I'm not sure I know what that would mean.
Hopefully this question makes sense; thanks for your assistance!
 A: It depends on how you define orthogonality, or, as OSE puts it in his comment, "Orthogonality is usually tested using some defined inner product."  I'll expand on this a bit.
In order to mathematically answer the question

Is direction A orthogonal to direction B?

we need a definition of the terms "direction" and "orthogonal."  The standard mathematical way to formalize the notion of direction is by using vectors.  
For example, imagine we are traveling along some curve $\gamma$ in the plane, and let's say we are at some point $x$ in the plane, then the direction of $\gamma$ at the point $x$ can be defined by a vector tangent to $\gamma$ at the point $x$.
In particular, let's say that $\gamma$ is just the $x$-axis, then a tangent vector to the $x$-axis at every point is just $(1,0)$ (or any positive scalar multiple of this), and this defines the direction of this line at every point.  Similarly, the direction of the $y$-axis at every point is defined by the vector $(0,1)$ (or any positive scalar multiple of this).
What about the notion of orthogonality?  Well since vectors define directions, we might be inclined to think that orthogonality of directions should be defined in terms of associating a number to each pair of vectors, and that when this number has a special value, we call these vectors (and therefore the directions they define) orthogonal.
In practice, that's exactly how it's done.  The association of a number to a given pair of vectors that tests for orthogonality is called an inner product, as mentioned in OSE's comment to your question.  Given any pair of vectors $u$ and $v$, it is common to see the inner product denoted by something like $u\cdot v$, or $\langle u,v\rangle$, or something similar, depending on the context.  Given an inner product, two vectors are said to be orthogonal with respect to that inner product provided their inner product is zero.
So let's take the example of directions in the plane.  The standard inner product on the plane, often referred to as the "dot product" is defined as follows:
\begin{align}
  (u_x, u_y)\cdot (v_x, v_y) = u_xu_y + v_xv_y
\end{align}
To test that two directions are orthogonal, we just need to take their inner product and verify that it's zero.  For example, the $x$- and $y$- directions are orthogonal since
\begin{align}
  (1,0)\cdot (0,1) = 1(0) + 0(1) = 0.
\end{align}
Now let's go back to the original question of time and space being orthogonal.  Let's restrict the discussion to $1+1$-dimensional spacetime with coordinates $(t,x)$ for simplicity.  The direction of the time axis is given by the unit vector $(1,0)$.  The direction of the space axis is given by the unit vector $(0,1)$.  Calling them orthogonal now depends on the inner product we specify.  
If we choose the inner product to be just like the dot product on the $x$-$y$ plane, namely if we choose
\begin{align}
  (u_t, u_x)\cdot (v_t, v_x) = u_tv_t + u_xv_x,
\end{align} 
then yes, time and space are orthogonal with respect to this product.
However, the physical interpretation and significance of applying this inner product to spacetime is murky.  The standard inner product on the plane is motivated by the fact that it comports with the usual notion of distance.  In particular, if two vectors are orthogonal with respect to this inner product, then the sum of the square of their lengths agrees with the independently defined notion of the Euclidean distance between their endpoints.  
In the case of spacetime, this notion of distance isn't particularly useful or appropriate.  There is, however a different notion of "distance" derived from a scalar product (which is not strictly speaking an inner product since it's not positive definite) defined by
\begin{align}
  (u_t, u_x)\cdot (v_t, v_x) = -u_tv_t + u_xv_x.
\end{align}
Unfortunately, since this scalar product is not an inner product, the notion of orthogonality is rather strained.  But if you insist on still calling vectors orthogonal if their scalar product with each other is zero, then the $x$ and $t$ directions are still orthogonal relative to this product.
A: 
How do we prove that any directions are orthogonal? [...] we can use the pythagorean theorem. 

This involves of course a definition of (how to measure or compare) "angle(s)" in the first place; such that one may comprehend statements about (distinct) angles being "equal" (or else: "not equal") for instance in Euclid's 4th axiom (on "right angles") or in Hilbert's "4th axiom of congruence" (concerning "angles" in general).
In a flat metric space, a suitable definition of an
"angle at point $B$, between directions $\vec{BA}$ and $\vec{BC}$", in terms of distance ratios between points $A$, $B$ and $C$, is:
$ \angle[ ABC ] := \text{ ArcSin } \! \! \! \left[ \frac{\Large 1}{\Large 2} \sqrt{ 2 + 2 \left(\frac{AC}{AB}\right)^2 + 2 \left(\frac{AC}{BC}\right)^2 - \left(\frac{AB}{BC}\right)^2 - \left(\frac{BC}{AB}\right)^2  - \left(\frac{AC}{AB}\right)^2 \left(\frac{AC}{BC}\right)^2 } \, \right] $.
More generally, one may define 
$ \angle[ ABC ] := \text{ Limit }{ \! \! }_{\{ F, G \}} \! \huge[ $
$ {\Large \{ } \left( \frac{BF}{AB} \right) \rightarrow 0, \left( \frac{BG}{BC} \right) \rightarrow 0, $
$ 2 + 2 \left(\frac{AB}{AF}\right)^2 + 2 \left(\frac{AB}{BF}\right)^2 - \left(\frac{AF}{BF}\right)^2 - \left(\frac{BF}{AF}\right)^2  - \left(\frac{AB}{AF}\right)^2 \left(\frac{AB}{BF}\right)^2 \rightarrow 0, $
$ 2 + 2 \left(\frac{BC}{BG}\right)^2 + 2 \left(\frac{BC}{GC}\right)^2 - \left(\frac{BG}{GC}\right)^2 - \left(\frac{GC}{BG}\right)^2  - \left(\frac{BC}{BG}\right)^2 \left(\frac{BC}{BC}\right)^2  \rightarrow 0 \Large\} \Large[ $
$ \text{ ArcSin } \! \! \! \left[ \frac{\Large 1}{\Large 2} \sqrt{ 2 + 2 \left(\frac{FG}{BF}\right)^2 + 2 \left(\frac{FG}{BG}\right)^2 - \left(\frac{BF}{BG}\right)^2 - \left(\frac{BG}{BF}\right)^2  - \left(\frac{FG}{BF}\right)^2 \left(\frac{FG}{BG}\right)^2 } \, \right] \Large] \huge] $.
In the case of a flat metric space, this lends itself to defining a
"scalar product between vectors $\vec{BA}$ and $\vec{BC}$" as
$ \langle \vec{BA}, \vec{BC} \rangle :=$
$ AB \, \, BC \, \, {\text{ Cos }} \! \! {\large[} \, \angle[ ABC ] \, {\large]} = $
$ AB \, \, BC \, \, \sqrt{ 1 - \frac{\Large 1}{\Large 4} \left(  2 + 2 \left( \frac{AC}{AB} \right)^2 + 2 \left(\frac{AC}{BC}\right)^2 - \left(\frac{AB}{BC}\right)^2 - \left(\frac{BC}{AB}\right)^2  - \left(\frac{AC}{AB}\right)^2 \left(\frac{AC}{BC}\right)^2 \right) } = $
$ AB \, \, BC \, \, \frac{\Large 1}{\Large 2} \sqrt{ 2 - 2 \left( \frac{AC}{AB} \right)^2 - 2 \left(\frac{AC}{BC}\right)^2 + \left(\frac{AB}{BC}\right)^2 + \left(\frac{BC}{AB}\right)^2 + \left(\frac{AC}{AB}\right)^2 \left(\frac{AC}{BC}\right)^2 } = $
$ AB \, \, BC \, \, \frac{\Large 1}{\Large 2} \sqrt{ 1 + \left(\frac{BC}{AB}\right)^2 - \left(\frac{AC}{AB}\right)^2 } \sqrt{ 1 + \left(\frac{AB}{BC}\right)^2 - \left(\frac{AC}{BC}\right)^2 } = $
$ \frac{\Large 1}{\Large 2} \left(  AB^2 + BC^2 - AC^2 \right) $,
where 
$ \langle \vec{BA}, \vec{BA} \rangle := AB^2$, and so on.  

Easy! My question is:
  can this method be used to prove that our concept of time is orthogonal to our concept of space?     

This method can be used for expressing relations between "spacelike" and "timelike" or "lightlike" vectors (in a suitably supplemented flat space). It can be made to match the assignment of "intervals" $s$: writing
$ \langle \vec{BA}, \vec{BA} \rangle = \langle \vec{AB}, \vec{AB} \rangle := s_{AB}^2$,
where
$ s_{AQ}^2 = 0 $ for vector $\vec{AQ}$ lightlike, 
$ s_{BQ}^2 \lt 0 $ for vector $\vec{BQ}$ timelike, and
$ s_{AB}^2 = -s_{BQ}^2 $ for the "magnitude square" of corresponding spacelike vector $\vec{AB}$, such that in this case indeed
$ \langle \vec{BA}, \vec{BQ} \rangle = \frac{\Large 1}{\Large 2} \left(  AB^2 + BQ^2 - AQ^2 \right) = \frac{\Large 1}{\Large 2} \left( s_{AB}^2 + s_{BQ}^2 + 0 \right) = 0 $
as is characteristic of orthogonal vectors.
A: From my answer to  Space time diagrams : Length contraction
Here's how Minkowski describes this...

From Minkowski's "Space and Time"...
We decompose any vector, such as that from O to x, y, z, t into four
components x, y, z, t. If the directions of two vectors are, respectively, that
of a radius vector OR from O to one of the surfaces ∓F = 1, and that of
a tangent RS at the point R on the same surface, the vectors are called
normal to each other. Accordingly,
$$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$
is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to
be normal to each other.

In other words,
locate the intersection of an observer's 4-velocity with the unit-hyperbola (the Minkowski circle) centered at the tail of the observer's 4-velocity.
The tangent line to that hyperbola is Minkowski-perpendicular to that observer's 4-velocity. That observer's x-axis is drawn through the tail of her 4-velocity, parallel to that tangent line.

The "intuition" to have is that
the tangent to the "circle" in that geometry
is
orthogonal to the radius vector.

To play with this idea, visit my
robphy's spacetime diagrammer for relativity v.8e-2021 (time upward)
https://www.desmos.com/calculator/emqe6uyzha 
Play with the E-slider to go from
Minkowski (E=+1), Galilean (E=0), and Euclidean (E=-1).
(See my answer in Meaning of simultaneity in special relativity for more info.)



A: "Can we show that time is orthogonal to space?"
Yes, but not by mathematics since the latter depends upon not differentiating as to how we 'register' space and time. Even (inertial) stationary rest mass illustrates this. You measure rest mass space position by the location it OCCUPIES relative to some reference inertial mass. BUT, you measure the time 'location' of this mass by a perception-measurement event. [The inertial rest mass has no defined time location; only events have such.]
This is Ontology 101; it does not require Mathematics 801. You measure space and time totally differently because rest mass resides in (occupies) only one dimension giving it no defined location in the other [time] dimension. Your time measure/location of the mass is orthogonal, via an EVENT.
Residing (occupying-existing) at a location and event (occurring) location are different.     [but no one pays attention to this :-( ]
