Sequentially Compact Metric Space is Lindelöf

Lemma
Let $M$ be a metric space.

Let $M$ be sequentially compact.

Then $M$ is also a Lindelöf space.

That is, from every open cover of $M$, it is possible to extract a countable subcover.

Proof
Let $M = \left({X, d}\right)$ be a sequentially compact metric space.

Take any open cover $C$ of $M$.

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

We have that a Sequentially Compact Metric Space is Second-Countable.

Thus, by definition, the topology of $M$ has a countable basis.

Let $\mathcal B$ be a countable basis for the topology induced by $d$ of $M$.

Let $x \in X$.

As $C$ covers $X$:


 * $\exists U_x \in C: x \in U_x$

As $\mathcal B$ is a basis:


 * $\exists B_x \in \mathcal B: x \in B_x \subseteq U_x$

Thus:
 * $(1): \quad \forall x \in X: \exists B_x \in \mathcal B: x \in B_x$

Consider the set $\Sigma := \left\{{B_x: x \in X}\right\}$.

$\Sigma$ is a subset of $\mathcal B$.

Hence $\Sigma$ is countable.

As $\Sigma$ contains every

From $(1)$, $\Sigma$ covers $X$.

By construction of $\Sigma$, every open set in $\Sigma$ is contained in some $U \in C$.

For each open set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$.

Let $\mathcal U$ be the set defined as:
 * $\mathcal U \left\{ {U_B: B \in \Sigma}\right\}$

As $\mathcal U$ does not have more sets than $\Sigma$, $\mathcal U$ is countable.

We have that:
 * $\forall B \in \Sigma: B \subseteq U_B$

Thus it follows that $\mathcal U$ covers $X$.

As:
 * $\forall U_B \in \mathcal U: U_B \in C$

$U$ is a countable subcover of $C$.

Thus a countable subcover has been obtained from $C$.

As $C$ is arbitrary, it follows that $M$ is a Lindelöf space.

Also see

 * Sequentially Compact Metric Space is Compact