Proof of The Restriction of a Metric to a Subset Induces the Subspace Topology

Claim 1. Let $a,b,c\in A$. Since $a,b,c$ also lie in $X$ and $d_A$ agrees with $d$ on $A\times A$, the four conditions in the definition of a metric for $d_A$ at $a,b,c$ are exactly the corresponding conditions for $d$ at $a,b,c$, which hold because $d$ is a metric on $X$. Hence $d_A$ is a metric on $A$.

For $a\in A$ and a real number $r>0$ write $B_{d}(a,r)$ and $B_{d_A}(a,r)$ for the open balls in $(X,d)$ and in $(A,d_A)$. Since $d_A$ agrees with $d$,

$$B_{d_A}(a,r)=\{b\in A: d(a,b)<r\}=A\cap B_{d}(a,r).$$

Claim 2, the implication from $\mathcal{T}_A$ to metric openness. Let $V=A\cap U$ with $U\in\mathcal{T}_d$, and let $a\in V$. Then $a\in U$, so by the definition of a subset open in $(X,d)$ there is a real number $r>0$ with $B_d(a,r)\subseteq U$. Hence

$$B_{d_A}(a,r)=A\cap B_d(a,r)\subseteq A\cap U=V .$$

As $a\in V$ was arbitrary, $V$ is open in $(A,d_A)$.

Claim 2, the implication from metric openness to $\mathcal{T}_A$. Let $V\subseteq A$ be open in $(A,d_A)$. For each $a\in V$ choose a real number $r_a>0$ with $B_{d_A}(a,r_a)\subseteq V$, and set

$$U=\bigcup_{a\in V}B_d(a,r_a),$$

the union of the family of open balls $B_d(a,r_a)$ indexed by $a\in V$. Each $B_d(a,r_a)$ belongs to $\mathcal{T}_d$ by Open Ball in a Metric Space is Open, and $\mathcal{T}_d$ is a topology by Metric Open Sets Form a Topology, so $U\in\mathcal{T}_d$ because a topology is closed under arbitrary unions. (If $V$ is empty, then $U$ is empty, which belongs to $\mathcal{T}_d$ for the same reason.)

We show $A\cap U=V$. If $b\in A\cap U$, then $b\in B_d(a,r_a)$ for some $a\in V$, so $b\in A\cap B_d(a,r_a)=B_{d_A}(a,r_a)\subseteq V$. Conversely, if $a\in V$, then $a\in A$ and $d(a,a)=0<r_a$ by condition 2 in the definition of a metric, so $a\in B_d(a,r_a)\subseteq U$. Hence $V=A\cap U\in\mathcal{T}_A$.

The two implications together prove claim 2.