Given that $\pi: X \to Y$ is a continuous onto map between compact metric spaces such that the fiber $\pi^{-1}(y)$ is a finite subset of $X$ for all $y$, is the map $y \mapsto \pi^{-1}(y)$ guaranteed to behave in a measurable way? For example, is it guaranteed that $y \mapsto |\pi^{-1}(y)|$ is Borel measurable, and that for each Borel measurable $A \subset X$, the subset $\{ y \in Y : A \cap \pi^{-1}(y) \neq \emptyset \}$ is Borel measurable?
For any concrete enough class of maps satisfying the assumption, there always seems to be a way to prove the conclusion just for that class, so I could not find any particular counter-example.
Remark 1. It is easy to think of a zig zag like map from the interval onto itself such that $y \mapsto |\pi^{-1}(y)|$ is neither upper nor lower semi-continuous. So the idea that $\{y \in Y: |\pi^{-1}(y)| \ge k\}$ might always be a closed subset does not work for establishing the partial conclusion that $y \mapsto |\pi^{-1}(y)|$ is measurable.
Remark 2. The measure theoretical counterpart (where measures are involved) has positive answer, but that implies nothing about the original problem. The measure theoretical counterpart is: given a measure preserving map $\pi: (X, \mu) \to (Y, \nu)$ between two Lebesgue spaces (i.e. standard probability space) such that the disintegration $\mu_y$ is supported on a finite subset of $X$ for $\nu$-a.e. $y$, then the map $y \mapsto \bar{\pi}(y)$ where $\bar{\pi}(y) := \{ x \in X: \mu_y(x) > 0 \}$, which is $\nu$-a.e. well defined (unlike the map $y \mapsto \pi^{-1}(y)$ which is not well defined), is a measurable map in some sense. For example, $\{ y \in Y : A \cap \bar{\pi}(y) \neq \emptyset \}$ is measurable for any measurable $A \subset X$. If $A$ in that expression runs over elements of appropriate finite partitions of $X$, then the map $y \mapsto \bar{\pi}(y)$ corresponds a map from $Y$ to the set of binary sequences, and the latter map is measurable, which we may take as the meaning of the former map being measurable.
(Update) Remark 3. The counterpart for the case where only Y (and not X) is a probability space also has a positive answer: if $\pi: X \to Y$ is a Borel-measurable (not necessarily onto) map from a Polish space $X$ to another Polish space $Y$ and if $\nu$ is a probability measure on $Y$, then it is easy to check that $\{y \in Y: |\pi^{-1}{y}| \ge k\}$ and $\{y \in Y: A \cap \pi^{-1}{y} \ne \emptyset \}$ are analytic sets and hence $\nu$-measurable. This might be useful for studying measures in $\pi_*^{-1}(\nu)$.