Sequentially Compact Metric Space is Compact

Theorem
A sequentially compact metric space is compact.

Proof using Lindelöf property
Let $$(X,d)$$ be a sequentially compact metric space. Our proof is based on the result that a sequentially compact metric space is Lindelöf, that is, every open cover of $$X$$ has a countable subcover.

Take any open cover $$C$$ of $$X$$, and extract from it a countable subcover $$\{U_1, U_2, \ldots \}$$.

Reasoning by contradiction, assume that there is no finite subcover of $$C$$. Then, for any natural $$n \geq 1$$, the family $$\{U_1,\ldots,U_n\}$$ does not cover $$X$$, so we can choose a point $$x_n \in X$$ such that
 * $$x_n \notin U_1 \cup \cdots \cup U_n.$$

In this way we have constructed an infinite sequence $$\{x_n\}_{n \geq 1}$$ of points of $$X$$.

As we are assuming $$X$$ is sequentially compact, this sequence has a subsequence which converges to some $$x \in X$$. But there is some $$U_m$$ such that $$x \in U_m$$ (as the $$U_i, i \geq 1$$, are a cover), and hence by one of the characterizations of convergence, there is an infinite of number of terms in the sequence $$\{x_i\}$$ which are contained in $$U_m$$.

This is a contradiction, as in the way we constructed our sequence, each $$U_n$$ can only contain a finite number of the terms ($$U_n$$ can contain only points $$x_i$$ with $$i < n$$).

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

Let $$\mathcal{U}$$ be any open cover of $$M$$.

By Lebesgue's Number Lemma, there exists a Lebesgue number for $$\mathcal{U}$$.

By Sequentially Compact Metric Space has Finite Net, there exists $$\left\{{x_1, x_2, \ldots, x_n}\right\}$$ which is a finite $\epsilon$-net for $$M$$, where $$\epsilon$$ is this same Lebesgue number.

Then $$N_{\epsilon} \left({x_i}\right)$$ is contained in some $$U_i \in \mathcal{U}$$ by definition of Lebesgue number.

Since $$M \subseteq \bigcup_{i=1}^n N_{\epsilon} \left({x_i}\right) \subseteq \bigcup_{i=1}^n U_i$$, we have a finite subcover $$\left\{{U_1, U_2, \ldots, U_n}\right\}$$ of $$\mathcal{U}$$ for $$M$$.

Hence the result.