Proof of A Compact Subset of a Metric Space is Sequentially Compact

Let $(x_m)_{m\in\mathbb{N}}$ be a sequence in $X$ with $x_m\in K$ for every $m\in\mathbb{N}$.

Since $K$ is compact in $(X,\mathcal{T}_d)$, Every Sequence in a Compact Subset of a Metric Space Has a Cluster Point There provides a point $x\in K$ that is a cluster point of $(x_m)_{m\in\mathbb{N}}$ in $(X,d)$.

By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(h_k)_{k\in\mathbb{N}}$ of real numbers with $0<h_k$ for every $k\in\mathbb{N}$ and with limit $0$.

Apply A Cluster Point of a Sequence in a Metric Space is the Limit of a Subsequence to the cluster point $x$ with $\varepsilon_k=h_k$. Claim 1 of that theorem yields a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$, in the sense of Subsequence of a Sequence in a Set, with $d(x_{n_k},x)<h_k$ for every $k\in\mathbb{N}$; and since $(h_k)_{k\in\mathbb{N}}$ has limit $0$, claim 2 of that theorem shows that the subsequence $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$.

Thus for every sequence in $X$ with values in $K$ there are a point $x\in K$ and a subsequence converging to $x$, which is exactly the requirement of Sequentially Compact Subset of a Metric Space. Hence $K$ is sequentially compact in $(X,d)$.