So, Here is a proof I came up for it,
Define relative position function $x_1$ is position of body-1 and $x_2$ of body-2:
$$x(t) = x_2 (t) - x_1 (t)$$
Then finding the maxima of this function leads us too
$v_2 = v_1$
But I can not make intuitive physical sense of this, how do we 'expect' this answer?
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{d\vec{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.
Intuitive physical interpretation: set one object as the frame of reference. If $v_2\not=v_1$, in the new frame of reference the other ball will have some velocity, either in the direction that makes two object closer or farther. In the first case, (closer direction), just wait some more time and they will get closer (therefore not the closest at this moment); in the second case, some time before they were closer (therefore not the closest at this moment). Hence the time they are the closest is when they have the same velocity.
Strict proof: use calculus and one can show both (local) closest or farthest moment would have their velocity the same.
