Proof that Lipschitz maps are uniformly continuous

Proof

We give an explicit $\delta$ for each $\varepsilon$.

1. Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. Let $f \colon X \to Y$ be Lipschitz with a constant $K \ge 0$.
2. So $d_Y(f(x), f(y)) \le K \, d_X(x, y)$ for all $x, y \in X$.
3. Let $\varepsilon > 0$ be real, and put $\delta = \varepsilon / (K + 1)$. Then $\delta > 0$, because $K + 1 \ge 1$.
4. Let $x, y \in X$ with $d_X(x, y) < \delta$.
5. By step 2 and $K \ge 0$, we get $d_Y(f(x), f(y)) \le K \, d_X(x, y) \le K \delta = \varepsilon K / (K + 1)$.
6. Since $K / (K + 1) < 1$ and $\varepsilon > 0$, we get $\varepsilon K / (K + 1) < \varepsilon$. So $d_Y(f(x), f(y)) < \varepsilon$.
7. The number $\delta$ depends only on $\varepsilon$ and $K$, not on $x$ or $y$.
8. So condition 1 of uniform continuity holds, and $f$ is uniformly continuous.