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

Suppose first that $K$ 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.

Conversely, suppose that $K$ is sequentially compact in $(X,d)$. Then $K$ is compact in $(X,\mathcal{T}_d)$ by A Sequentially Compact Subset of a Metric Space is Compact.

The two implications together give the stated equivalence.