Metric Subspace Sequentially Compact in Itself iff Closed

Theorem

Let $M$ be a metric space.

Let $C \subseteq M$ be a subspace of $M$ which is sequentially compact in $M$.

Then $C$ is sequentially compact in itself iff $C$ is closed in $M$.

Proof

Follows directly from Closure of Subset of Metric Space by Convergent Sequence.

$\blacksquare$