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 $A = \emptyset$ the conclusion of claim 3 holds by its second alternative, so assume $A \ne \emptyset$; 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 $A = V_0 \cup \dots \cup V_n$.

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 $A = V_0 \cup \dots \cup V_n = (U_{i_0} \cap A) \cup \dots \cup (U_{i_n} \cap A) \subseteq U_{i_0} \cup \dots \cup U_{i_n}$, 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 $A = \emptyset$ 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 $A \subseteq W_0 \cup \dots \cup W_m$.

Putting $G_j := W_j \cap A$ for $j \le m$ gives members of $\mathcal{G}$ with $A = (W_0 \cap A) \cup \dots \cup (W_m \cap A) = G_0 \cup \dots \cup G_m$, 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.