Real Analysis Folland, Corollary 2.2 This follows from Proposition 2.1. 
Corollary 2.2: If $X$ and $Y$ are metric (or topological) spaces, every continuous $f:X\rightarrow Y$ is $(\mathcal{B}_X,\mathcal{B}_Y,)$-measurable.
proof (similar to proof in book): Suppose we have a open set $U$ and we have $U\subset Y$. Since $f:X\rightarrow Y$ we can state that $f$ is continuous if and only if $f^{-1}(U)$ is open in $X$. Then since, $\mathcal{B}_X$ and $\mathcal{B}_Y$ are $\sigma$-algebras generated by the family of open or closed sets in the metric space $X$ and $Y$ respectively, then clearly $f$ is $(\mathcal{B}_X,\mathcal{B}_Y,)$-measurable.
I am not sure if this makes sense or not or if the proof is not rigorous enough, any suggestions is greatly appreciated.  
 A: Your answer is more or less correct, though you should clarify how exactly the result of Proposition 2.1 is used. Worded differently, Proposition 2.1 states that the function $f$ (as defined in that proposition) is measurable if and only if the preimage of every set that generates $N$ is a measurable set in $Y$ (i.e. it is an element of $M$).
To that end, let $\mathcal{O}_Y$ be the family of open sets in $Y$, then we know that this collection of sets generates the Borel $\sigma$-algebra $\mathcal{B}_Y$. Given any open set $U_Y \in \mathcal{O}_Y$, we know that $f^{-1}(U_Y)$ is an open set of $X$, by continuity of $f$. Since the open sets of $X$ generate the Borel $\sigma$-algebra $\mathcal{B}_X$ and $\sigma$-algebras always contain their generating sets, we have that $f^{-1}(U_Y) \in \mathcal{B}_X$ for every $U_Y \in \mathcal{O}_Y$ i.e. the preimage of every generating set of $\mathcal{B}_Y$ is measurable in $X$. By Proposition 2.1, we conclude that $f$ is a measurable function.
A: You have shown that if $U\subset Y$ is open then $f^{-1}(U)$ is an element of $\mathcal B_X$. Why is it "clear" (to you) that the same is true if $U$ is merely an element of $\mathcal B_Y$?
