Let $(X, d)$ be a complete separable metric space, and let $Y$ be a finite subset of $X$. The associated Voronoi diagram is the collection of sets $\{R_y\}_{y \in Y}$ where
$$R_y :=\bigl\{x \in X : d(x, y) = \min\limits_{y' \in Y}\,d(x, y')\bigr\}\,.$$
Question: Does there exist a collection $\{S_y\}_{y \in Y}$ of disjoint Borel-measurable subsets of $X$ satisfying $\bigcup_{y \in Y}S_y = X$ and $S_y \subseteq R_y$ for all $y \in Y$?
On
The sets $$R_y=\bigcap_{y'\in Y}\left\{x\in X: d(x,y')-d(x,y)\geq 0\right\}$$ are closed being intersections of preimages of a closed set under a continuous function. Write $Y=\{y_1,\ldots,y_n\}$ and define $$ S_{y_{k}}=R_{y_{k}}\backslash \bigcup_{j=1}^{k-1} R_{y_j} $$ with the agreement that $S_{y_1}=R_{y_1}$. Then $S_{y_k}$ form a disjoint Borel partition of $X$ with $S_{y_k}\subseteq R_{y_k}$ for every $k$.
Let $n$ be the cardinality of $Y$, and let $p$ be a bijection of $Y$ and $\{1,2,\ldots,n\}$. Define $f(x):=\min\{p(y): x\in R_y, y\in Y\}$ and $S_y:=R_y\cap\{x\in X: f(x)=p(y)\}=R_y\cap f^{-1}(\{p(y)\})$. Clearly $S_y\subset R_y$ for each $y\in Y$, and $\cup_{y\in Y}S_y=X$. Also, each $R_y$ is closed, and so the function $f_y$ defined by $$ f_y(x):=\cases{p(y),&$x\in R_y$, \cr n+1,& $x\notin R_y$,\cr} $$ is lower semi-continuous, hence Borel measurable. But $f$ is the pointwise minimum of $\{f_y:y\in Y\}$, and so $f$ is also Borel measurable. It follows that each $S_y$ is a Borel set.