**Lemma**: Any closed subset $S$ of a [[Compact topological spaces|compact topological space]] $(X, \mathcal{T})$ is compact. **Proof**: Let $\mathcal{U}$ be a cover of $S$ by open subsets of $X$. Then $\mathcal{U} \cup\{X \backslash S\}$ is an open cover of $X$, so has a finite subcover; elements of this subcover (removing $X \backslash S$ if it is included) provide a finite open subcover of $S$.