A Sequentially Compact Subset of a Metric Space is Totally Bounded

Let $(X,d)$ be a metric space, and let $K\subseteq X$ be sequentially compact in $(X,d)$.

Then for every real number $\varepsilon>0$ there exists a finite subset $F\subseteq K$ such that

$$K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon),$$

where $B_d(a,\varepsilon)$ is the open ball in $X$ with center $a$ and radius $\varepsilon$, and where the union over the empty set is empty. In particular $K$ is totally bounded in $(X,d)$.