Solution: Note that given any $x,y\in \mathbb{R} $, we have \[ \vert f(x) - f(y) \vert \leq |x - y|^2, \] which, in particular, implies that $f$ is a continuous function. So for any $x,h \in \mathbb{R} $, we have \begin{align*} & \vert f(x + h) - f(x) \vert \leq \vert (x + h) - x \vert ^2 \\ \implies & \frac{\vert f(x + h) - f(x) \vert }{\vert h \vert } \leq \vert h \vert \\ \implies & 0 \leq \frac{\vert f(x + h) - f(x) \vert }{\vert h \vert } \leq \vert h \vert. \end{align*} By Sandwich's theorem, we get \[ \lim_{h \to 0} \frac{\vert f(x + h) - f(x) \vert }{\vert h \vert } = 0, \] thus $f$ is differentiable at $x$ with $f'(x) = 0$ for $x\in \mathbb{R} $. Therefore, $f$ is constant.