The relation between subspace topology and sub-$\sigma$-algebra I'm trying to prove this result. Could you verify if my attempt is fine?

Let $(X, \tau_X)$ be a topological space and $\mathcal B(X)$ its Borel $\sigma$-algebra generated by $\tau_X$. Let $Y$ be a subset of $X$, $\tau_Y$ the subspace topology on $Y$, and $\mathcal B(Y)$ the Borel $\sigma$-algebra on $Y$ generated by $\tau_Y$. Then $\mathcal B(Y) = \{B \cap Y \mid B \in \mathcal B(X)\}$.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.
 A: I think your proof is fine, though I'm not 100% sure. I have a proof below that is a little slower.

We prove that $\mathscr{B}(Y) = \mathscr{B}(X) \cap Y := \bigl\{ V \cap Y \colon V \in \mathscr{B}(X)\bigr\}$ in two steps.

*

*First, note that $\bigl\{ V \cap Y \colon V \in \mathscr{B}(X)\bigr\}$ is a $\sigma$-algebra (the trace $\sigma$-algebra) containing $\tau_Y$, because $\tau_X \subseteq \mathscr{B}(X)$. Since $\mathscr{B}(Y)$ is the smallest $\sigma$-algebra containing $\tau_Y$, we must have $\mathscr{B}(Y) \subseteq \bigl\{ V \cap Y \colon V \in \mathscr{B}(X)\bigr\}$.


*Second, note that the inclusion map $i \colon Y \to X$ given by $y \mapsto y$ is continuous. Since it is continuous, and we are using the Borel $\sigma$-algebra on both sides, the inclusion map is also measurable. Using the definition of measurability for a function, we can take a set $B \in \mathscr{B}(X)$ and know that $i^{-1}(B) \in \mathscr{B}(Y)$. But by the definition of the inclusion map, we know that $i^{-1}(B) = B \cap Y$. This implies that $\bigl\{ V \cap Y \colon V \in \mathscr{B}(X) \bigr\} \subseteq \mathscr{B}(Y)$.
A: Consider the canonical injection $i: Y \to X, y \mapsto y$. Then $\tau_Y$ is the initial topology with $i$ as the reference map. We have $\tau_Y = i^{-1} (\tau_X)$. By this result, we have $i^{-1} (\sigma(\tau_X)) = \sigma (i^{-1} (\tau_X))$, so
$$
\sigma(\tau_Y) = i^{-1} (\sigma(\tau_X)) = i^{-1} (\mathcal B(X)) =  \{i^{-1}(B) \mid B \in \mathcal B(X)\} = \{B \cap Y \mid B \in \mathcal B(X)\}.
$$
This completes the proof. It's worth noticing that if $Y \in \mathcal B(X)$ then $\mathcal B(Y) \subset \mathcal B(X)$.
