$x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$ I'm trying to prove below equivalence. Could you verify if my understanding is correct?

Let $(X, \tau)$ be a metric space, $A \subseteq X$, and $\tau_A$ the subspace topology of $A$. Let $a\in A$ and $(x_d)_{d \in D}$ is a net in $A$. Then $x_d \to a$ in $\tau$ if and only if $x_d \to a$ in $\tau_A$.

First, we need a lemma.

Lemma: $U'$ is a neighborhood (nbh) of $a$ in $A$ if and only if there is a nbh $U$ of $a$ in $X$ such that $U' =U \cap A$.
Proof:

*

*If $U'$ is a nbh of $a$ in $A$, then there is $O' \in \tau_A$ such that $a \in O' \subseteq U' \subseteq A$. Also, there is $O \in \tau$ such that $O' = O \cap A$. Let $U := O \cup U'$. Then $a \in O \subseteq U$ and thus $U$ is a nbh of $a$ in $X$. We have $U \cap A = (O \cup U') \cap A = (O \cap A)\ \cup (U' \cap A) = O' \cup  U' = U'$.

*If $U$ is a nbh of $a$ in $X$, then there is $O \in \tau$ such that $a\in O \subseteq U$. So $a \in (O \cap A) \subseteq (U\cap A)$. Clearly, $O' := O\cap A \in \tau_A$ and thus $U' := U \cap A$ is a nbh of $a$ in $A$.


My attempt:
Let $x_d \to a$ in $\tau$ and $U'$ is a nbh of $a$ in $A$. By lemma, $U' = U\cap A$ for some nbh $U$ of $a$ in $X$. There is $d' \in D$ such that $x_d \in U$ and thus $x_d \in U'$ for all $d \ge d'$. Hence $x_d \to a$ in $\tau_A$.
Let $x_d \to a$ in $\tau_A$ and $U$ is a nbh of $a$ in $X$. By lemma, $U' := U\cap A$ is a nbh of $a$ in $A$. There is $d' \in D$ such that $x_d \in U'$ and thus $x_d \in U$ for all $d \ge d'$. Hence $x_d \to a$ 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.
