Closed subset of a compact set is compact

Closed subset of compact is compact: Let $(X,d)$ be a metric space , let $K\subseteq X$ be compact , and let $F\subseteq K$ be closed in $X$ (equivalently, closed in the subspace $K$). Then $F$ is compact.

This permanence property is used constantly: once compactness is established, it automatically applies to all closed substructures.

Proof sketch: Let $\{U_\alpha\}$ be an open cover of $F$ in $X$. Then $\{U_\alpha\}\cup\{X\setminus F\}$ is an open cover of $K$. By compactness of $K$, there is a finite subcover. Removing $X\setminus F$ leaves a finite subcover of $F$.