First of all, you can't prove anything in physics (or any natural science by the way). What you can do is either build models which are in agreement with a certain theory or with a new one you're willing to create. If the predictions arriving from you model agree with the experiments, it's a good model, otherwise, it's a bad one and won't be used. Nobody proved a formula for kinetic energy, people simply recognized that a certain expression showed up so often it was worth giving it a name. That said, there are some things to consider regarding its form.
You can find a very good explanation regarding the form of the kinetic energy expression in Landau's classical mechanics book, which you should check if you can. When we talk about classical mechanics, some things are tacitly assumed , uniformity of time and space included. We must assume that there is a function which contains all mechanical information of the system, the Lagrangian. Let's think about the simple case of ONE free particle. It can be described in generalized coordinates, but let's do it in Cartesian terms, so L is, in general, a function of the spatial coordinates, time, and velocity (which are here independent). But time and space are postulated to be uniform, so, if our function is to be consistent with the said uniformity of space and time, it can't contain time and position explicitly (why, if there's no force acting on my particle I can't expect it to change simply because time passed, nor should I expect that one particular point in space is at all different of the others). Also, space is considered isotropic, so I must be able to rotate my entire system without changing it's physical properties. Some requirements are already clear: my function can't, for instance, be a function of v^k , with k odd, 'cause otherwise a rotation of my system, which changes the sign of velocity, would change my Lagrangian, and that's not allowed. My function must depend only on the absolute value of velocity. Now comes the point when me must choose which function, among the infinite that respect those requirements. Let's take L= L(v^2) for now. Yes, we coude have choosen differently, but I'll come to it later.
Now, given this assumption, let's change our reference frame, and choose one that moves with a infinitely small velocity \delta in relation to our previous fame. By Galileo's principle, this change can't alter the physical properties of my system, so the Lagrangian in this new frame must obey: L'=L(v'^2)=L([v+\delta]²). A Taylor expansion around \delta of the rightmost expression ( remember that \delta can be made as small as desired) , renders : L(v'^2)=L(v^2) + \frac{\partial L}{\partial v^2}2V\delta , where V\equiv \delta - V' (I assumed up there that our new reference frame moved with a small velocity in relation to the previous one, so here I'm just repeating myself in mathematical terms!)
We have two different Lagrangians representing the same system . Since the laws of physics are still the same, the equations of movement must still be equal, dispate this difference. The only way this could happen is if they differ only by a total time derivative (if you didn't know it, examine the Euler-Lagrange and you will see). Hence:
\frac{dL(v^2)}{dt} \equiv \frac{\partial L}{\partial v^2} \frac{\partial L}{\partial t} = \frac{\partial L}{\partial v^2}2V\delta \to \frac{\partial L}{\partial t} = 2V\delta \implies L= \alpha v^2 , where \alpha is the constant which arrives from integration. Giving it the value (m/2), we arrived at the desired expression.
Now, could I choose a different value for the constant? Mathematically, yes. But since our question was simply whether the usual formula was wrong, I've proved that no, it isn't. It's consistent with the axioms (Notice that I used nothing but the very basic axioms of mechanics, I didn't even specify the form of the Lagrangian at first.), so there's zero possibility that it's wrong. Any formula that can be proved right by the same method would be mathematically right. If you want to call \alpha by a different name, you wil arrive in a consistent, but completely different (and probably less convenient) formalism for mechanics. Any different choices would imply a new form for Newton's second law, a new definition of what we call mass, and so on. If, instead of v2, you wished to use v4, for example, you could be able to come up with a self consistent formalism in which the kinetic energy isn't given by the usual expression, but if you want any of the expressions to agree with Newton's formulation and to our basic, intuitive concepts, you would need to make exact the choices made above.
