Proof of Compact Subset Criterion via Open Covers in the Ambient Space

Write $\mathcal{T}_A$ for the subspace topology on $A$, so that $\mathcal{T}_A=\{A\cap U : U\in\mathcal{T}\}$. By The Subspace Topology is a Topology the pair $(A,\mathcal{T}_A)$ is a topological space, so by Compact Topological Space and Compact Subset statement 1 says exactly the following: for every set $I$ and every family of subsets of $A$ $(V_i)_{i\in I}$ with $V_i\in\mathcal{T}_A$ for every $i\in I$ and $A\subseteq\bigcup_{i\in I}V_i$, there is a finite subset $J\subseteq I$ with $A\subseteq\bigcup_{j\in J}V_j$.

Statement 1 implies statement 2. Assume statement 1, and let $(U_i)_{i\in I}$ be an open cover of $A$ in $X$. For $i\in I$ put $V_i=A\cap U_i$. Then $V_i\in\mathcal{T}_A$ for every $i\in I$. Moreover, if $x\in A$, then $x\in U_i$ for some $i\in I$, hence $x\in A\cap U_i=V_i$; therefore $A\subseteq\bigcup_{i\in I}V_i$. By statement 1 there is a finite subset $J\subseteq I$ with $A\subseteq\bigcup_{j\in J}V_j$. Since $V_j\subseteq U_j$ for every $j$, this gives $A\subseteq\bigcup_{j\in J}U_j$, which is statement 2.

Statement 2 implies statement 1. Assume statement 2. Let $I$ be a set and let $(V_i)_{i\in I}$ be a family of subsets of $A$ with $V_i\in\mathcal{T}_A$ for every $i\in I$ and $A\subseteq\bigcup_{i\in I}V_i$. Put

and for $p=(i,U)\in P$ set $W_p=U$. Because $V_i\in\mathcal{T}_A$, for every $i\in I$ there is at least one $U\in\mathcal{T}$ with $V_i=A\cap U$, so every $i\in I$ occurs as the first coordinate of some element of $P$. Note that $P$ collects all admissible pairs, so no choice of a distinguished $U$ for each $i$ is made.

The family $(W_p)_{p\in P}$ is a family of subsets of $X$ with $W_p\in\mathcal{T}$ for every $p\in P$. It covers $A$: if $x\in A$, then $x\in V_i$ for some $i\in I$, and choosing an element $(i,U)\in P$ with first coordinate $i$ we get $x\in V_i=A\cap U\subseteq U=W_{(i,U)}$. Hence $(W_p)_{p\in P}$ is an open cover of $A$ in $X$, and statement 2 provides a finite subset $Q\subseteq P$ with

Let

be the set of first coordinates of elements of $Q$. We check that $J$ is finite. If $Q=\emptyset$, then $J=\emptyset$, which is finite by Finite Set. If $Q\ne\emptyset$, then, being finite and nonempty, $Q$ has $m$ elements for some natural number $m$, and the map $q:Q\to J$ sending $(i,U)$ to $i$ is surjective by the definition of $J$; hence $J$ is finite by claim 4 of Basic Properties of Finite Sets.

Finally, $A\subseteq\bigcup_{j\in J}V_j$. Indeed, let $x\in A$. Then $x\in W_p$ for some $p=(i,U)\in Q$, so $x\in U$, and since also $x\in A$ we get $x\in A\cap U=V_i$ with $i\in J$.

Thus every family of subspace-open subsets of $A$ covering $A$ admits a finite subcover indexed by a finite subset of the index set, which is statement 1.