Why do two objects at distance of closest approach have equal velocity?

If you fix two trajectories $vec{x}_1(t)$ and $vec{x}_2(t)$, and want to determine when they are closer, you should find the minimum of the function begin{equation} d = |vec{x}_1(t) - vec{x}_2(t)|, end{equation} and not of $vec{x}(t) = vec{x}_1(t) - vec{x}_2(t)$, otherwise you obtain two different results. Let us take a two dimensional example, such as two trajectories which do not meet: begin{align} vec{x}_1(t) &= (at + b, 0) &&a,b>0 vec{x}_2(t) &= (0,c) && c>0 end{align} Here $d = sqrt{(at + b)^2 + c^2}$ and by differentiation you find that the minimum is for $t = -frac{b}{a}$ as expected. If you instead applied your condition, you would have found $a=0, c=0$, which is meaningless, since you have fixed the trajectories.

Please note that the minimum distance might not be obtained through differentiation! Take a simple one-dimensional example: begin{align} x_1(t) &:= at + b, &&a,b>0 x_2(t) &:= c,&&c,d>0 end{align} The time when the objects are closer is of course the time when $x_1(t) = x_2(t)$, but that is exactly the point where $d$ is not differentiable. The one-dimensional example is somehow trivial because the two objects always meet (but maybe in the past!), so you can never use differentiation. I was able to use it in my first example because the two trajectories never met.

In order to minimize a non-differentiable function, such as $d$, you should look for its stationary points, and then verify if the absolute value of the function in the non-differentiable point is lesser or not from the stationary ones.

A physical interpretation of your condition, could come if you have not taken $vec{x}_1(t)$ and $vec{x}_2(t)$ as fixed, but rather as generic unkown functions. Then, the condition begin{equation} frac{dvec{x} (t)}{dt} = 0, end{equation} would mean imposing that $vec{x}$ does not change in time. Among the generic functions, you select the ones where the relative position of the two objects does not change in time: of course it means that the difference in their velocity is 0, meaning that $vec{v}_1(t) = vec{v}_2(t)$, which is exactly the condition you find.