Example 4 Let $f(x)$ be a function defined on the set of non-negative real numbers, $f(0)=0$, and for any $x \geqslant y \geqslant 0$, we have
$$
|f(x)-f(y)| \leqslant(x-y) f(x) \text {. }
$$

Find the expression for $f(x)$.
Prove by mathematical induction: For any positive integer $n$, when $\frac{n-1}{2} \leqslant x<\frac{n}{2}$, we have $f(x)=0$.
When $n=1$, take $0 \leqslant x<\frac{1}{2}, y=0$ in the condition, we get $|f(x)| \leqslant x f(x) \leqslant x|f(x)| \leqslant \frac{1}{2}|f(x)|$.
Thus, $|f(x)|=0$, i.e., $f(x)=0$ holds on $\left[0, \frac{1}{2}\right)$.
Assume the proposition holds for $n=k$.
Then, when $n=k+1$, take $x, y$ such that
$\frac{k}{2} \leqslant x<\frac{k+1}{2}, y=x-\frac{1}{2}$, at this time, $\frac{k-1}{2} \leqslant y<\frac{k}{2}$.
By the induction hypothesis, $f(y)=0$.
Substitute into the known condition to get $|f(x)| \leqslant \frac{1}{2} f(x)$.
Similarly, we conclude $f(x)=0$.
Thus, the proposition holds for $n=k+1$.
Notice that, when $n$ takes all positive integers, $x \in \left[\frac{n-1}{2}, \frac{n}{2}\right)$ covers all non-negative real numbers.
By mathematical induction, $f(x) \equiv 0$.
[Note] This problem applies mathematical induction in the context of real numbers, which can be seen as a variation of mathematical induction: unlike proving propositions about positive integers, because the set of positive real numbers is "uncountable," it is necessary to verify infinitely many starting points. The approach in this problem is to decompose the interval $[0,+\infty)$ into countably many intervals $\left[\frac{n-1}{2}, \frac{n}{2}\right)$, and unify the verification process for infinitely many starting points by discussing the initial interval, then set the "step size" to $\frac{1}{2}$ for the proof.