Lipschitz maps are uniformly continuous

Statement

Let $X$ and $Y$ be metric spaces and let $f \colon X \to Y$ be a Lipschitz map. Then $f$ is uniformly continuous.

Remarks

Proof: let $K > 0$ be a Lipschitz constant of $f$. For $\varepsilon > 0$, take $\delta = \varepsilon / K$. If $d(x, y) < \delta$, then $d(f(x), f(y)) \le K d(x, y) < \varepsilon$. If $f$ is Lipschitz with $K = 0$, then $f$ is constant. The converse fails: $x \mapsto \sqrt{x}$ on $[0, 1]$ is uniformly continuous, but it is not Lipschitz. In particular, each Lipschitz map is continuous.

Explanations

L1 human

For a Lipschitz map, you can give an explicit $\delta$ for each $\varepsilon$: divide $\varepsilon$ by the constant $K$. This $\delta$ does not depend on the point, so the map is uniformly continuous.