Proof of Compactness in a Subspace Agrees with Compactness in the Ambient Space

Claim 1. Let $V\in\mathcal{T}_B$. By the definition of $\mathcal{T}_B$ there is $U\in\mathcal{T}$ with $V=B\cap U$, and then

$$A\cap V=A\cap(B\cap U)=(A\cap B)\cap U=A\cap U,$$

where the last equality holds because $A\subseteq B$ gives $A\cap B=A$. Hence every set of the form $A\cap V$ with $V\in\mathcal{T}_B$ is of the form $A\cap U$ with $U\in\mathcal{T}$.

Conversely, let $U\in\mathcal{T}$ and put $V=B\cap U$. Then $V\in\mathcal{T}_B$, and by the computation just made $A\cap V=A\cap U$. Hence every set of the form $A\cap U$ with $U\in\mathcal{T}$ is of the form $A\cap V$ with $V\in\mathcal{T}_B$. The two collections therefore coincide, which is claim 1.

Claim 2. By The Subspace Topology is a Topology the pair $(B,\mathcal{T}_B)$ is a topological space, so the phrase compact in $B$ is meaningful and says, by that definition, that $A$ is a compact topological space when equipped with the subspace topology inherited from $(B,\mathcal{T}_B)$, namely with the collection $\{A\cap V : V\in\mathcal{T}_B\}$. Likewise, $A$ is compact in $X$ exactly when $A$ is a compact topological space when equipped with the collection $\{A\cap U : U\in\mathcal{T}\}$.

By claim 1 these two collections of subsets of $A$ are equal, so the two assertions concern one and the same topological space and are therefore equivalent.