The best proof of Clairaut's theorem on equality of mixed partials

This is the best proof of Clairaut's theorem on equality of mixed partials that I have ever thought up. And I bet it's the best proof of the theorem you'll ever see!

I'll prove the 2-variable version, but the generalization to $n$-variables is obvious.

Clairaut's Theorem: Suppose $f$ is a real-valued function of two variables $x,y$ and $f(x,y)$ is defined on an open subset $U$ of $\mathbb{R}^2$. Suppose further that both the second-order mixed partial derivatives $\partial_x\partial_y f(x,y)$ and $\partial_y\partial_x f(x,y)$ exist and are continuous on $U$. Then, we have: $$\partial_x\partial_y f = \partial_y\partial_x f$$ on all of $U$.



Proof: Fix some $(x_0,y_0) \in U$. Consider the "discrete approximation" function
$$F(h_x, h_y) = \frac{1}{h_x h_y} (f(x_0 + h_x, y_0+h_y) - f(x_0, y_0 + h_y) - f(x_0 + h_x, y_0) + f(x_0, y_0))$$
where $h_x, h_y > 0$ are small enough such that the rectangle $R(h_x, h_y) = [x_0, x_0+h_x]\times [y_0, y_0+h_y]$ lies entirely within $U$.

It's important to understand the meaning of $F(h_x, h_y)$. It can be viewed as a discrete approximation of either $\partial_x\partial_y f(x_0, y_0)$:
$$\partial_x\partial_y f(x_0, y_0)
\approx \frac{1}{h_x}(\partial_y f(x_0+h_x, y_0) - \partial_y f (x_0, y_0))
\approx \frac{1}{h_x}\left[\frac{1}{h_y}\left(f(x_0 + h_x, y_0+h_y) - f(x_0 + h_x, y_0 )\right) - (f(x_0 , y_0+ h_y) - f(x_0, y_0))\right] = F(h_x, h_y)$$
or similarly for $f_{yx}(x_0, y_0)$.

Then apply Rolle's Theorem twice, for any small $h_x, h_y > 0$, this approximation is actually exact in a sense: We have
$$F(h_x, h_y) = \frac{1}{h_x}(\partial_y f(x_0+h_x, \xi_y) - \partial_y f (x_0, \xi_y)) = \partial_x\partial_y f(\xi_x, \xi_y)$$
for some $\xi_y \in [y_0, y_0 + h_y], \xi_x \in [x_0, x_0 + h_x]$. The precise values of $\xi_x, \xi_y$ aren't important, the only thing to know is that they are functions of $h_x, h_y$, and $(\xi_x, \xi_y) \in R(h_x, h_y)$

Similarly, $F(h_x, h_y) = \partial_y\partial_x f(\xi_x', \xi_y')$ for some  $(\xi_x', \xi_y') \in R(h_x, h_y)$.

So, for all such rectangles $R(h_x, h_y)\in U$, there exists two points $(\xi_x, \xi_y) , (\xi_x', \xi_y') \in R(h_x, h_y)$, such that $\partial_x\partial_y f(\xi_x, \xi_y) = \partial_y\partial_x f(\xi_x', \xi_y')$.

Now take the limit $(h_x, h_y) \to (0, 0)$, so the rectangle shrinks to $(x_0, y_0)$. By continuity of  $\partial_x\partial_y f$ and $\partial_y\partial_x f$, we have $\partial_x\partial_y f(x_0, y_0) = \partial_y\partial_x f(x_0, y_0)$.