Lemma 14.2.6. Let $X$ be a topological space and $Y \subset X$ be a subspace, then the Borel $\sigma$-algebra on $Y$ coincides with the $\sigma$-algebra on $Y$ induced by $\cb_{X}$.
Proof. Since $\bracsn{A \in \cb_X|A \cap Y \in \cb_Y}$ is a $\sigma$-algebra that contains all open sets in $X$, $\cb_{Y}$ contains the induced $\sigma$-algebra.
On the other hand, the induced $\sigma$-algebra contains all open sets in $Y$, so it contains $\cb_{Y}$.
Therefore the two $\sigma$-algebras coincide.$\square$