Why is time-order invariant in timelike interval? Why do two observers measure the same order of events if we are inside the light cone?
(e.g. if $ds^2 > 0$ time-order is preserved according to the classical mechanics book I am reading, but it doesn't give any proof of this) I assume there is some simple geometrical argument that I am missing. And why do two observers measure possible different order of events if we are outside the light cone?
 A: For a geometrical argument, you're looking for basically what Ron posted. But you can also argue this one mathematically: as you may know, the difference between two spacetime events is represented by a time difference $\Delta t$ and a spatial difference $\Delta x$. Under a Lorentz boost, these quantities transform like this:
$$\begin{align}c\Delta t' &= \gamma(c\Delta t - \beta\Delta x) \\
\Delta x' &= \gamma(\Delta x - \beta c\Delta t)\end{align}$$
Now, the spacetime interval is $\Delta s^2 = c^2\Delta t^2 - \Delta x^2$. For a timelike interval, $\Delta s^2 > 0$, this means $c\Delta t > \Delta x$, assuming that both differences are positive (and you can always arrange for that to be the case). Using the Lorentz boost equations, you can see that in this case, $c\Delta t'$ has to be positive. So for two events separated by a timelike interval, if one observer (in the unprimed reference frame) sees event 2 later than event 1, any other observer (in the primed reference frame) will also see event 2 later than event 1.
On the other hand, suppose you have a spacelike interval, $\Delta s^2 < 0$. In this case, $\Delta x > c\Delta t$, so it is possible to get $c\Delta t' < 0$ for a specific velocity (namely $\beta > \frac{c\Delta t}{\Delta x}$). So if one observer (in the unprimed reference frame) sees event 2 later than event 1, it's still possible for another observer (in the primed reference frame) to see them in the reverse order.
A: The circles in geometry are the curves with
$$ x^2 + y^2 = C $$
In relativity, the analog of circles are hyperbolas:
$$ t^2 - x^2 - y^2 - z^2 = C $$
These curves, unlike circles, are disconnected hyperbolas. For any x,y,z, and positive C, there are two solutions for t, positive and negative, and they are never closer than 2C in t. The two branches of the hyperbola go up in time, and down, and define the forward and backward branch of the hyperbola.
Much as a rotation takes points around a circle, a lorentz transformation takes points along the hyperbola. Those Lorentz transformations which rotate the point continuously cannot move points from the upper hyperbola to the lower hyperbola.
Any timelike interval is either in the forward or backward hyperbola, and is either strictly to the future, or to the past. Null intervals too, by continuity.
A: To get a feel for Lorentz 'rotations' in spacetime, you might want to have a look at this GIF: 

Notice the events outside the light cone to move up and down in response to the accelerations of the reference frame, and as a result, these can end up at both sides of the 'now' of the observer at the origin. This is not the case for events within the light cone. It is these latter events that can have an influence on the observer at the origin. 
A: I will restate the answers by Ron and DavidZ in slightly different words. The interval is $ds^2=dt^ 2-dx^2$ in natural units. Thus, $dt^ 2=ds^ 2+dx^ 2$. Thus, if $ds^ 2$ is positive (i.e., the interval is timelike) then no transformation that leaves $ds^ 2$ invariant can make $dt^ 2$ zero as $dx^2$ remains non-negative by definition. Thus, no transformation which is connected to the identity transformation can change its sign because in order to do so it would have to first make $dt$ zero which is forbidden as $dt^2$ cannot go to zero. 
