Compact subsets of a metric space are bounded

Statement

Let $(X, d)$ be a metric space and let $K$ be a compact subset of $X$. Then $K$ is bounded.

Remarks

Proof: if $K$ is empty, it is bounded. Otherwise fix $x \in K$. The open balls $B(x, n)$ for $n \ge 1$ cover $K$, so finitely many of them do. The largest of these balls contains $K$. A compact subset of a metric space is also closed, because metric spaces are Hausdorff. The converse fails in general: a closed bounded set need not be compact. With the metric $\min(|x - y|, 1)$, the real line is closed and bounded, but not compact.

Explanations

L1 human

A compact set can be covered by finitely many balls about one point, and then it fits in the largest of them.