Sequential compactness equals compactness (metric spaces)

Sequential compactness equals compactness: Let $(X,d)$ be a metric space and $K\subseteq X$. Then $K$ is compact (every open cover has a finite subcover) if and only if $K$ is sequentially compact (every sequence in $K$ has a convergent subsequence with limit in $K$).

This equivalence is special to metric (first countable) spaces and makes compactness usable via sequences, which is often the most practical viewpoint in analysis.

Proof sketch (optional): Compact $\Rightarrow$ sequentially compact: if no convergent subsequence exists, one can build an infinite collection of separated points and then an open cover with no finite subcover. Sequentially compact $\Rightarrow$ compact: if an open cover has no finite subcover, construct a sequence avoiding larger and larger finite subcollections and show it contradicts sequential compactness.