Epsilon-net characterisation of total boundedness

Statement

Let $(X, d)$ be a metric space and let $S \subseteq X$. Then $S$ is totally bounded if and only if for each real $\varepsilon > 0$ there is a finite set $F \subseteq S$ with $$S \subseteq \bigcup_{y \in F} B(y, \varepsilon).$$

Remarks

Such a finite set $F$ is an $\varepsilon$-net of $S$. Here $B(y, \varepsilon)$ is an open ball. Proof: an $\varepsilon$-net gives a cover by balls of diameter at most $2\varepsilon$. Conversely, from a cover by sets $A_k$ of diameter less than $\varepsilon$, choose one point of $S$ in each $A_k$ that meets $S$. The centres can be taken in $S$ or in $X$. Both choices give the same notion.

Explanations

L1 human

For each $\varepsilon$, you can place finitely many points in $S$ such that each point of $S$ is within $\varepsilon$ of one of them. For $[0, 1]$ and $\varepsilon = 0.1$, the points $0, 0.1, 0.2, \ldots, 1$ do the job.