There's an interesting property that if $(X,\mathcal{T})$ and $(Y,\mathcal{S})$ are topological spaces, then the Borel $\sigma$-algebra of $X\times Y$ with the product topology includes the product $\sigma$-algebra of the Borel $\sigma$-algebras of the respective spaces, and if the spaces are actually second countable, then these two $\sigma$-algebras coincide.
Does this still hold if one of the spaces is only known to be a metric space? Let's say $(X,\mathcal{T})$ is second countable, but $(Y,d)$ is only known to be a metric space. Does it still hold that the Borel $\sigma$-algebra on $X\times Y$ is the same as the product $\sigma$-algebra of the Borel $\sigma$-algebras on $X$ and $Y$?
I tried taking an arbitrary open set $A$ in $X\times Y$. Since $X$ is second countable, there exists a countable basis $\{U_n\}_{n\in\mathbb{N}}$. So for any $n$ and $\epsilon>0$, set $V_{n,\epsilon}=\{y\in Y\mid\text{ for some }\delta>0, U_n\times B_{\epsilon+\delta}(y)\subset A\}.$ [Here $B_r(s)$ is the ball centered at $s$ of radius $r$.] I was then trying to show that $A$ can be written as an arbitrary union of sets of form $U_n\times V_{n,\epsilon}$, with $\epsilon$ possibly varying, to prove it. Is there a way to flesh out this idea?