Sequentially Compact Metric Space is Compact/Proof 1

Theorem
A sequentially compact metric space is compact.

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

From Sequentially Compact Metric Space is Lindelöf, every open cover of $A$ has a countable subcover.

Let $C$ be an open cover $A$.

Extract from it a countable subcover $\left\{{U_1, U_2, \ldots}\right\}$.

Aiming for a contradiction, suppose that there exists no finite subcover of $C$.

Then, for any $n \in \N_{\ge 1}$, the set $\left\{{U_1, \ldots, U_n}\right\}$ does not cover $A$.

Hence it is possible to choose $x_n \in A$ such that:
 * $x_n \notin U_1 \cup \cdots \cup U_n$.

Thus we construct an infinite sequence $\left\{{x_n}\right\}_{n \ge 1}$ of points of $A$.

By assumption $A$ is sequentially compact

Thus $\left\{{x_n}\right\}_{n \ge 1}$ has a subsequence which converges to some $x \in A$.

But because the $U_i: i \ge 1$ forms a cover for $A$, there exists some $U_m$ such that $x \in U_m$.

Hence by one of the characterizations of convergence, there is an infinite number of terms in the sequence $\left\langle{x_i}\right\rangle$ which are contained in $U_m$.

But from the method of construction of $\left\langle{x_i}\right\rangle$, each $U_n$ can contain only points $x_i$ with $i < n$.

That is, each $U_n$ can only contain a finite number of the terms of $\left\langle{x_i}\right\rangle$.

This is a contradiction.

Thus the supposition that there exists no finite subcover of $C$ was false.

Hence the result.