Q:

Computing relative velocities from angular velocity

Consider two people on a disk of radius R, distances $r_1$ > $r_2$ from the center. 
Suppose the disk rotates at angular velocity $\vec{\omega}$. The velocities of the two people relative to the ground are $\vec{\omega}$ x $\vec{r_1}$ and  $\vec{\omega}$ x $\vec{r_2}$ respectively. If you then calculated the velocity of person 2 from the frame of reference of person 1, you would get $\vec{\omega}$ x $\vec{r_2}$ - $\vec{\omega}$ x $\vec{r_1} \ne 0$. But, if two people are on a carousel they should measure each other's velocity as 0.  

A:

Every point on a rotating rigid body is moving with different tangential velocity according to the rule you stated
$$\vec{v} = \vec{ \omega } \times \vec{r} $$
Yet the distance $\ell_{AB}$ between two arbitrary points A and B remains the same. This can only occur of the difference in velocities $\vec{v}_A = \vec{\omega} \times \vec{r}_A$ and $\vec{v}_B=\vec{\omega} \times \vec{r}_B$ is only along the perpendicular to the line connecting A and B.
It is sufficient to prove that $$(\vec{v}_A - \vec{v}_B) \cdot (\vec{r}_A - \vec{r}_B) =0$$ NOTE: $\cdot$ is the dot product and $\times$ is the cross product
Proof
Expand out the velocity terms
$$ ( \vec{\omega} \times \vec{r}_A -\vec{\omega} \times \vec{r}_B) \cdot ( \vec{r}_A-\vec{r}_B) =\left( \vec{\omega} \times ( \vec{r}_A-\vec{r}_B) \right) \cdot ( \vec{r}_A-\vec{r}_B) =0 $$
There is vector identity for triple products that states $\left(\vec{a} \times \vec{b} \right) \cdot \vec{b} =0 $

