Compact iff complete and totally bounded

Statement

Let $(X, d)$ be a metric space and let $K \subseteq X$. Then $K$ is compact if and only if $K$, with the subspace metric, is complete and $K$ is totally bounded.

Remarks

Proof idea of the converse: in a totally bounded set, a diagonal argument gives a Cauchy subsequence of each sequence. Completeness gives a limit. Thus $K$ is sequentially compact, and so compact. In a complete space $X$, a subset is compact if and only if it is closed and totally bounded. The Heine–Borel theorem follows, because bounded subsets of $\mathbb{R}^n$ are totally bounded.

Explanations

L1 human

Compactness in a metric space splits into two parts: no holes, and finitely many small pieces for each size. The interval $(0, 1)$ is totally bounded but has holes at its ends. The real line has no holes but is not totally bounded.