Sequentially Compact Metric Space is Lindelöf

Lemma
A sequentially compact metric space is Lindelöf.

That is, from every cover one can extract a countable subcover.

Proof
Let $(X,d)$ be a metric space, and take any open cover $C$.

We need to find a countable subset of $C$ which still covers $X$.

We will use the result that the topology of a sequentially compact metric space has a countable base.

So, take a countable basis $\mathcal B$ for the topology $(X,d)$.

For each $x \in X$:
 * Take $U_x \in C$ such that $x \in U_x$ (which can be done, as $C$ covers all of $X$).
 * Take $B_x \in \mathcal B$ such that $x \in B_x \subseteq U_x$ (which can be done because $\mathcal B$ is a base).

Then, consider the family $\Sigma := \{B_x \mid x \in X \}$.

This family is a subset of $\mathcal B$, and hence is countable (notice there must be many repetitions here if the space $X$ is large; $B_x$ may be the same for many points $x$).

Also, $\Sigma$ covers $X$ (it contains every point $x \in X$, as $x \in B_x$).

Now, for each (open) set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$ (this can be done because every open set in $\Sigma$ is contained in some $U \in C$, by construction).

This family:
 * $\{ U_B \mid B \in \Sigma \}$

is what we want:


 * It is countable (as it does not have more sets than $\Sigma$).
 * It is an open cover of $X$, as it covers more than $\Sigma$ (for each $B \in \Sigma$, $B \subseteq U_B$), and $\Sigma$ already covers $X$.
 * It is a subcover of $C$, as we chose each $U_B$ among the sets in $C$.

This proves that $X$ is Lindelöf.

Also see
This result is part of a possible proof of a more general result, namely, that a sequentially compact metric space is compact (i.e., from every cover one can extract a finite subcover).