Proof that Lipschitz functions are uniformly continuous

Proof

1. Since $f$ is Lipschitz, there is a real $K \ge 0$ with $\abs{f(x) - f(y)} \le K \abs{x - y}$ for all $x, y \in D$.
2. Let $\varepsilon > 0$. Put $\delta = \frac{\varepsilon}{K + 1}$. Then $\delta > 0$, since $K + 1 > 0$.
3. Let $x, y \in D$ with $\abs{x - y} < \delta$. By step 1, $\abs{f(x) - f(y)} \le K \abs{x - y} \le K \delta$.
4. Since $K < K + 1$ and $\delta > 0$, we have $K \delta < (K + 1) \delta = \varepsilon$. So $\abs{f(x) - f(y)} < \varepsilon$.
5. The number $\delta$ depends only on $\varepsilon$, not on $x$ or $y$. So $f$ is uniformly continuous.