How do we prove the existence of the instantaneous axis of rotation? Euler's theorem of rotation states that any rigid body motion with one point fixed is equivalent to a rotation about some axis passing through that fixed point. Now it is often said that Euler's theorem of rotations implies the existence of an instantaneous axis of rotation.  My question is, how can we prove that the instantaneous axis of rotation exists?
Consider a rigid body which is undergoing some motion with one point fixed.  For any given times $t_1$ and $t_2$, let $\hat{u}(t_1,t_2)$ denote the unit vector parallel to the axis of the rotation that's equivalent to the motion of the the rigid body between time $t_1$ and time $t_2$.  Then how can we prove that the limit of $\hat{u}(t_1,t_2)$ as $t_2$ goes to $t_1$ exists?
Also, having proved the limit does exist, how can we prove that unit vector we get points in the direction of the angular velocity vector of the rigid body?
This would all be much easier to prove if the angular velocity vector were the derivative of the angular displacement vector, but things don't work out so smoothly because rotations don't commute; see this journal paper for details.
EDIT: I just posted a follow-up question here.
 A: Find the locus of points with no motion on a rigid body, if point A is fixed.
Without loss of generality place a coordinate system on A. Rigid body motion is defined if an arbitrary point maintains constant distance from A $$d = \sqrt{x^2+y^2+z^2}$$
This is found by setting $\frac{{\rm d}d}{{\rm d}t}=0$ which by the chain rule (with time derivative of $\frac{1}{2}d^2$) implies $$ x \frac{{\rm d}x}{{\rm d}t} +  y \frac{{\rm d}y}{{\rm d}t}+ z \frac{{\rm d}z}{{\rm d}t} = 0$$
In vector form the above is $$\vec{r} \cdot \vec{v} =0$$ with $\vec{r}=(x,y,z)$ and $\vec{v} = \frac{{\rm d}\vec{r}}{{\rm d}t}$.
So the velocity field has to be perpendicular to the location vector. An obvious solution to the above is
$$ \vec{v} = \vec{\omega} \times \vec{r}$$
$$ \vec{r} \cdot (\vec{\omega} \times \vec{r}) \equiv 0 $$
So we have established that vector a velocity field of the form $\vec{v} = \vec{\omega} \times \vec{r}$ describes rigid body motion. This also imposes the limitation that $\vec{\omega}$ is constant throughout the body because otherwise the above expression wouldn't be zero. Try it.
Now let's look at the points parallel to $\vec{\omega}$ with $\vec{r} = \lambda \,\vec{\omega}$. The velocities are
$$\vec{v}_\parallel =  \vec{\omega} \times (\vec{r}) = \vec{\omega} \times \lambda \,\vec{\omega} \equiv 0$$
Now let's look at points perpendicular to the direction of $\vec{\omega}$ with a distance $d$
$$\vec{r} = \lambda \vec{\omega} + d \vec{n}$$ with the properties $\| \vec{n} \|=1$ and $\vec{n}\cdot\vec{\omega}=0$
$$\vec{v} = \vec{\omega} \times (\vec{r}) = \vec{\omega} \times ( \lambda \,\vec{\omega} + d \,\vec{n}) = d\,(\vec{\omega} \times  \vec{n})$$
which is a vector perpendicular to both $\vec{\omega}$ and $\vec{n}$. This implies a tangential (hoop) direction for the velocity with the magnitude increasing linearly with distance.
We call this motion rotation.

Here is an interesting fact. If at some arbitrary point located at $\vec{r}$ the velocity vector is $\vec{v}$ and the body is rotating with $\vec{\omega}$ then the instant axis of rotation is located at
$$\vec{q} = \vec{r} + \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2}$$
The proof is that $\vec{v} = \vec{\omega} \times (\vec{r}-\vec{q}) \rightarrow$
$$\require{cancel}
\begin{align} 
  \vec{v} & = \vec{\omega} \times \left(\vec{r}-\left(\vec{r} + \frac{\vec{\omega} \times \vec{v}}{\| \vec{\omega} \|^2}\right) \right) \\
& =  -\frac{\vec{\omega} \times(\vec{\omega} \times \vec{v})}{\| \vec{\omega} \|^2} = -\frac{ \vec{\omega}( \vec{\omega} \cdot \vec{v} ) - \vec{v} (\vec{\omega} \cdot \vec{\omega})}{\| \vec{\omega} \|^2} =  \frac{ \vec{v} \| \vec{\omega}\|^2}{\| \vec{\omega} \|^2} - \frac{\vec{\omega}(\vec{\omega}\cdot\vec{v})}{\| \vec{\omega} \|^2} \\
\vec{v} &= \vec{v} -\frac{\vec{\omega}(\cancel{\vec{\omega}\cdot\vec{v}})}{\| \vec{\omega} \|^2} & \vec{v} \equiv \vec{v}
\end{align}$$
