Proof of Compactness and Sequential Compactness Agree for Nonempty Subsets of a Metric Space

If $K$ is compact in $(X,\mathcal{T}_d)$, then $K$ is sequentially compact in $(X,d)$ by A Compact Subset of a Metric Space is Sequentially Compact; this direction does not use the hypothesis that $K$ is nonempty.

Conversely, if $K$ is sequentially compact in $(X,d)$, then, $K$ being nonempty, A Sequentially Compact Subset of a Metric Space is Compact shows that $K$ is compact in $(X,\mathcal{T}_d)$.