Compactness implies total boundedness

Compactness implies total boundedness: Let $(X,d)$ be a metric space and let $K\subseteq X$ be compact. Then $K$ is totally bounded: for every $\varepsilon>0$ there exists a finite set $F\subseteq K$ such that

K\subseteq \bigcup_{x\in F} B(x,\varepsilon),

where $B(x,\varepsilon)$ denotes the open ball.

Remarks

This is one half of the standard metric characterization compact iff complete and totally bounded, complementing compactness implies completeness and closely related to the existence of finite epsilon-nets.