Let $0 \le d \le \frac{1}{4}$. Then we can find $x_1$ such that $x_1^2-x_1=d$ and $x_2$ such that $x_2^2-x_2=-d$. It follows that $f(x+d) \le f(x) \le f(x+d)$ for all $x$ and hence $f(x)=f(x+d)$ for all $x$. So $f$ is periodic with each period in $\left[0,\frac{1}{4}\right]$ and hence constant.

Very nice and simple proof!