Sequentially Compact Metric Subspace is Sequentially Compact in Itself iff Closed
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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 if and only if $C$ is closed in $M$.
Proof
Follows directly from Closure of Subset of Metric Space by Convergent Sequence.
$\blacksquare$