Continuous image of compact set is compact

Continuous image of compact set is compact: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces , let $K\subseteq X$ be compact , and let $f:X\to Y$ be continuous . Then $f(K)\subseteq Y$ is compact.

This is one of the most important permanence properties of compactness and is used to prove the extreme value theorem and many existence statements.

Proof sketch (optional): If $\{V_\alpha\}$ is an open cover of $f(K)$, then $\{f^{-1}(V_\alpha)\}$ is an open cover of $K$. Compactness of $K$ gives a finite subcover, whose images cover $f(K)$.