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

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $A \subseteq X$ and let $(A, \mathcal{T}_A)$ be the subspace (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). Then:

1. Compactness read in the ambient space. $A$ is a compact subset of $X$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), that is $(A, \mathcal{T}_A)$ is a compact space, if and only if for every family $\mathcal{U} \subseteq \mathcal{T}$ with $A \subseteq \bigcup \mathcal{U}$ there are $n \in \mathbb{N}$ and $U_0, \dots, U_n \in \mathcal{U}$ with $A \subseteq U_0 \cup \dots \cup U_n$, or else $A = \varnothing$.
2. The same in indexed form. $A$ is a compact subset of $X$ if and only if for every set $I$ and every family $(U_i)_{i \in I}$ of open subsets of $X$ with $A \subseteq \bigcup_{i \in I} U_i$ there are $n \in \mathbb{N}$ and indices $i_0, \dots, i_n \in I$ with $A \subseteq U_{i_0} \cup \dots \cup U_{i_n}$, or else $A = \varnothing$.

Claim 2 is the form used by almost every later proof on this page, because a cover is usually produced by a rule that attaches an open set to each point or to each index, and a set of open sets forgets that rule. No choice principle is used anywhere below; the one place a selection is made is over a finite index set, and Every natural-number-indexed list of nonempty sets has a choice function on its family of values is a theorem of ZF.

Why the ambient reading needed a proof at all. A subset $A$ of $X$ carries two candidate notions of open cover: families of sets open in $(A, \mathcal{T}_A)$, and families of sets open in $X$ whose union contains $A$. The trace description of the subspace topology is what turns one into the other, and it shows that compactness can be checked using ambient open sets for this fixed induced topology. Another ambient is guaranteed to give the same answer when it induces the same topology on $A$; if the induced topology changes, the answer may change. Every later item on this page that covers a subset by ambient open sets is using claim 1 or claim 2, and says so.

The traces do not remember their sources. A single relatively open $V$ is usually the trace of many different ambient open sets, and that is exactly why step 3.1 has to recover indices at all. Recovering infinitely many at once would be a choice principle; recovering finitely many is not, and the proof is arranged so that only finitely many are ever needed.

The metric statement of the same fact is 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, whose claims 2 and 3 are claims 1 and 2 above with the open subsets of a metric space in place of the members of an abstract topology. Its proof carries an extra first claim, that relative openness in a metric subspace is a trace, which here is the definition of the subspace topology and so needs no argument. Neither statement is used in the proof of the other; that the two agree is 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.