Compactness and sequential compactness agree in metric spaces

Statement

Let $(X, d)$ be a metric space and let $K \subseteq X$. Then $K$ is compact if and only if $K$ is sequentially compact.

Remarks

One proof of the hard direction uses the Lebesgue number lemma and the fact that a sequentially compact metric space is totally bounded. Both conditions are also equivalent to: $K$ is complete and totally bounded. This is the statement compact iff complete and totally bounded. Without a metric, the two notions differ. For example, the product $\{0, 1\}^{[0, 1]}$ is compact but not sequentially compact.

Explanations

L1 human

In a metric space, you can test compactness with sequences. A set is compact exactly when each sequence in it has a subsequence with a limit in the set. This test is often easier than the test with open covers.