A Closed Totally Bounded Subset of a Complete Metric Space is Compact

In the setting of The Real Numbers: Standing Notation and Background, let $(X,d)$ be a complete metric space, let $\mathcal{T}_{d}$ be the collection of subsets of $X$ that are open in $(X,d)$, a topology on $X$ by Metric Open Sets Form a Topology, and let $K\subseteq X$ be closed in $(X,\mathcal{T}_{d})$ and totally bounded in $(X,d)$. Then the following hold.

1. (Sequential compactness)¶ $K$ is sequentially compact in $(X,d)$: every sequence in $K$ has a subsequence converging in $(X,d)$ to a point of $K$.

2. (Compactness)¶ $K$ is compact in $(X,\mathcal{T}_{d})$.