How do I find all functions $F$ with $F(x_1) − F(x_2) le (x_1 − x_2)^2$ for all $x_1, x_2$?

The intended solution seems to be something like the following:

Fix some $x in mathbb{R}$. By assumption, for any $h in mathbb{R}$ (in particular, for any very small value of $h$), taking $x_1 = x+h$ and $x_2 = x$, we get

$$ F(x+h) - F(x) le big( (x-h) - x big)^2 implies frac{|F(x+h) - F(x)|}{|h|} le |h|. $$

Take $h$ to zero, do a little algebra (the limit passes into the absolute value, since the absolute value function is continuous at $0$), and apply the Squeeze Theorem to get

$$ Bigglvert underbrace{lim_{hto 0} frac{F(x+h) - F(x)}{h}}_{=F'(x),text{ if it exists}} Biggrvert le lim_{hto 0} |h| = 0.$$

This implies that $F$ is differentiable at $x$, and that $F'(x) = 0$. But $x$ was chosen arbitrarily, so $F$ is differentiable everywhere and $F' equiv 0$. Therefore $F$ is a constant function.

Exchanging $x_1$ with $x_2$ in the original inequality shows $F(x_1)-F(x_2)$ is bound by $pm(x_1-x_2)^2$, i.e. $left|frac{F(x_1)-F(x_2)}{x_1-x_2}right|le|x_1-x_2|$. This proves the two-sided derivative is $0$.

But you actually don't need derivatives to solve the problem. Since $|F(x)-F(0)|le x^2$ for all $x$, $|F(x)-F(0)|lelim_{ntoinfty}nleft(frac{x}{n}right)^2=lim_{ntoinfty}frac{x^2}{n}=0$ by the triangle inequality.