$x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$

Lose the metric space assumption. It's true in general and the proof you came up with did not use any metric specific facts as all.

Just say $(X,\tau)$ is a space and $\tau_A := \{O \cap A\mid O \in \tau\}$ is its subspace topology when $A \subseteq X$. (don't mention the definition of the subspace topology in the statement though, it's supposed to be known beforehand anyway).

Next, the lemma is redundant as convergence of nets is defined only in terms of open sets: $$(x_d)_{d \in D} \to p \iff \forall O \in \tau: ( p \in O) \to \left(\exists d_0 \in D: \forall d \in D: (d \ge d_0) \to (x_d \in U)\right)$$

Then the first paragraph of the proof becomes: let $(x_d)_{d \in D} \to a$ in $(X,\tau)$. Let $U' \in \tau_A$ such that $a \in U'$. By definition we can write $U'=U \cap A$ for some $U \in \tau$. Then there is some $d' \in D$ so that for all $d \ge d'$ we have $x_d \in U$ and as $x_d \in A$ by assumption, we in fact have $x_d \in A \cap U = U'$ for those $d$. Hence $(x_d)_{d \in D} \to a$ in $(X,\tau_A)$ as well.

The second paragraph needs similar minor edits.

The ideas are OK.