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

Suppose $U$ is open in $(X,d)$ and put $V:=U\cap A$; for $a\in V$ we have $a\in U$, so there is $r>0$ with $B_X(a,r)\subseteq U$, whence $B_A(a,r)=B_X(a,r)\cap A\subseteq U\cap A=V$, and $V$ is open in $(A,d_A)$.

Conversely let $V\subseteq A$ be open in $(A,d_A)$, and let $\mathcal{W}:=\{B_X(a,r):a\in V,r>0,B_X(a,r)\cap A\subseteq V\}$, a family cut out by a property of the pair $(a,r)$ and not by any selection; put $U:=\bigcup \mathcal{W}$.

$U$ is open in $(X,d)$, being a union of open balls.

$U\cap A\subseteq V$, since every member $B_X(a,r)$ of $\mathcal{W}$ satisfies $B_X(a,r)\cap A\subseteq V$ by the defining condition of $\mathcal{W}$.

$V\subseteq U\cap A$: given $a\in V$, openness of $V$ in $(A,d_A)$ gives $r>0$ with $B_A(a,r)\subseteq V$, that is $B_X(a,r)\cap A\subseteq V$, so $B_X(a,r)\in \mathcal{W}$ and $a\in B_X(a,r)\subseteq U$, while $a\in A$ as well.

Claim 1 is proved: by steps 2.2 and 2.3 an open $V\subseteq A$ equals $U\cap A$ with $U$ open in $X$ by step 2.1, and conversely every such trace is open in $(A,d_A)$ by step 1.1.

For claim 3, suppose first that $(A,d_A)$ is compact, let $I$ be a set and let $(U_i{)}_{i\in I}$ be open subsets of $X$ with $A\subseteq {\bigcup }_{i\in I}{U}_i$; then each $U_i\cap A$ is open in $(A,d_A)$ and $\mathcal{V}:=\{U_i\cap A:i\in I\}$ is a family of open subsets of $A$ whose union is $A$.

If $A=\mathrm{\varnothing }$ the conclusion of claim 3 holds by its second alternative, so assume $A\mathrm{\ne }\mathrm{\varnothing }$; then $\mathcal{V}$ is an open cover of $(A,d_A)$, and compactness yields $n\in \mathbb{N}$ and $V_0,\dots ,V_n\in \mathcal{V}$ with $A=V_0\cup \cdots \cup V_n$.

For each $j\le n$ the set $S_j:=\{i\in I:U_i\cap A=V_j\}$ is nonempty by the definition of $\mathcal{V}$, and $j\mapsto S_j$ is a function with domain the natural number $\sigma (n)$, so a choice function for its values supplies $i_0,\dots ,i_n\in I$ with $U_{i_j}\cap A=V_j$ for every $j\le n$.

Hence $A=V_0\cup \cdots \cup V_n=(U_{i_0}\cap A)\cup \cdots \cup (U_{i_n}\cap A)\subseteq U_{i_0}\cup \cdots \cup U_{i_n}$, which is the conclusion of claim 3 for the family $(U_i{)}_{i\in I}$, so the forward implication of claim 3 holds.

The converse of claim 3 remains, the forward implication having been settled at step 7.1; so assume the displayed condition, let $\mathcal{G}$ be a family of sets open in $(A,d_A)$ with union $A$, and put $\mathcal{U}:=\{U\subseteq X:U\text{ is open in }(X,d)\text{ and }U\cap A\in \mathcal{G}\}$, again a family cut out by a property, indexed by itself.

$A\subseteq \bigcup \mathcal{U}$: given $a\in A$ there is $G\in \mathcal{G}$ with $a\in G$, and by claim 1 there is $U$ open in $X$ with $U\cap A=G$; that $U$ lies in $\mathcal{U}$ and contains $a$.

If $A=\mathrm{\varnothing }$ the empty subfamily of $\mathcal{G}$ covers $A$; otherwise the assumed condition applied to the family $\mathcal{U}$ indexed by itself gives $m\in \mathbb{N}$ and $W_0,\dots ,W_m\in \mathcal{U}$ with $A\subseteq W_0\cup \cdots \cup W_m$.

Putting $G_j:=W_j\cap A$ for $j\le m$ gives members of $\mathcal{G}$ with $A=(W_0\cap A)\cup \cdots \cup (W_m\cap A)=G_0\cup \cdots \cup G_m$, so $\mathcal{G}$ has a finite subcover and $(A,d_A)$ is compact.

Claim 3 is proved by steps 7.1 and 11.1, and claim 2 is the special case of claim 3 in which $I=\mathcal{U}$ is a family of open subsets of $X$ and $U_i:=i$, the conclusion of claim 3 then naming members of $\mathcal{U}$ itself.