For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $\mathcal{T}_d$ be its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), so that $(X, \mathcal{T}_d)$ is a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and is metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Then:

1. $(X,d)$ is a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space) if and only if $(X, \mathcal{T}_d)$ is a compact topological space (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).
2. For every $A \subseteq X$: $A$ is a compact subset of the metric space $(X,d)$ if and only if $A$ is a compact subset of the topological space $(X, \mathcal{T}_d)$, the two readings of "compact subset" being the metric subspace $(A, d_A)$ (Isometry, isometric embedding, and the subspace metric on a subset) and the topological subspace $(A, (\mathcal{T}_d)_A)$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Nothing here is a coincidence and nothing is transported. The open-cover condition of Open cover, subcover, compact metric space, and compact subset of a metric space quantifies over families of subsets open in $(X,d)$, and by The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement those are exactly the members of $\mathcal{T}_d$; so the two conditions are not merely equivalent, they are the same condition written twice. No choice principle is used.

What this theorem buys, and why it is stated so early on the page. Every theorem proved on compactness-in-metric-spaces about compact metric spaces and their compact subsets is, by this theorem, a theorem about metrizable topological spaces and their compact subsets, once a metric inducing the topology has been named. Heine-Borel in $\mathbb{R}^n$ (Heine-Borel in $\mathbb{R}^n$: with the Euclidean metric a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), the extreme value theorem (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value) and the equivalence of the four compactness conditions for metric spaces (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice) are all available in that form and are used below without being reproved.

It also forbids a second notion. Since the metric development already fixed the intrinsic reading of "compact subset" (Open cover, subcover, compact metric space, and compact subset of a metric space) and this page fixes the same reading (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), claim 2 says that the phrase means one thing throughout the library, whichever of the two developments a reader arrives from. Had either page taken the ambient reading as its definition the phrase would have meant two things, and the agreement would have had to be proved through A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it and A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it rather than directly.