Compact Subspace of Metric Space is Sequentially Compact in Itself

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $C \subseteq M$ be a subspace of $M$ such that $C$ is compact.

Then $C$ is sequentially compact in itself.

Proof
Let $C \subseteq M$ be compact.

Let $\left \langle {x_n} \right \rangle$ be a sequence in $C$.

Let $S$ be the range of $\left \langle {x_n} \right \rangle$.

Thus $S \subseteq C$ and $S$ may be either finite or infinite.

Finite Range
Let $S$ be finite.

Then at least one $x \in S$ must be repeated infinitely often in $\left \langle {x_n} \right \rangle$.

Its occurrences form a subsequence converging to $x$.

Infinite Range
Let $S$ be infinite.

From Subsequence of Sequence in Metric Space with Limit, it is enough to show that $S$ has a limit point in $C$.

Aiming for a contradiction, suppose $S$ has no limit point in $C$.

Then for each $x \in C$, there exists $\epsilon \in \R_{>0}$ such that the open $\epsilon$-ball $B_\epsilon \left({x}\right)$ contains no $y \in S: y \ne x$.

That is:
 * $B_\epsilon \left({x}\right) \cap S = \left\{{x}\right\}$

or
 * $B_\epsilon \left({x}\right) \cap S = \varnothing$

Because $C$ is compact, the open cover $\mathcal B := \left\{{B_\epsilon \left({x}\right): x \in C}\right\}$ has a finite subcover.

But each $B_\epsilon \left({x}\right) \in \mathcal B$ contains at most one point of $S$.

Therefore the union of any finite subset of $\mathcal B$ contains only finitely many points of $S$.

So no finite subset of $\mathcal B$ covers $S$, let alone $C$.

Thus $S$ must have a limit point in $C$.