I think the answer is no. Consider the following subsets of the Euclidean plane.

Let $X_0 = \{(p,q)\in (\mathbb R\setminus \mathbb Q)\times \mathbb R: 0\le p \le 1, 0 \le q \le 1 \}$, $X_1 = \{(p,q)\in ( \mathbb Q)\times \mathbb R: 0\le p \le 1, -1 \le q \le 0 \}$,

Let $X=X_0\cup X_1$, $Y=[0,1]$, and let $f$ be the projection.

I think the answer is no. Consider the following subsets of the Euclidean plane.

Let $X_0 = \{(p,q)\in (\mathbb R\setminus \mathbb Q)\times \mathbb R: 0\le p \le 1, 0 \le q \le 1 \}$, $X_1 = \{(p,q)\in ( \mathbb Q)\times \mathbb R: 0\le p \le 1, -1 \le q \le 0 \}$,

Let $X=X_0\cup X_1$, $Y=[0,1]$, and let $f$ be the projection.

(ADDED: $X$ is not Polish, as $X$ contains a closed subset homeomorphic to $\mathbb Q$. Closed subsets of Polish spaces are Polish, but $\mathbb Q$ is not.)