Let $X$ and $Y$ be bounded complete separable metric spaces. Let $C = 2^\omega$ be Cantor space with its standard metric. All product spaces are taken to have the max metric.

Let $F, G \subseteq X\times Y$ be closed sets such that $\inf \{d(a,b) : a \in F, b \in G \} > 0$. For each $y\in Y$, let $F_y = \{ x\in X:(x,y) \in F \}$ and $G_y = \{ x \in X: (x,y) \in G\}$.

Does there always exist a set $Q \subseteq Y\times C$ and a function $f:Q \rightarrow 2^X$ such that

- for every $y\in Y$ there exists a $c\in C$ such that $(y,c) \in Q$,
- for every $q\in Q$, $f(q)$ is a closed subset of $X$,
- $f$ is uniformly continuous in the Hausdorff metric, and
- for every $(y,c) = q \in Q$, $F_y \subseteq f(q)$ and $f(q) \cap G_y = \varnothing$?

What if we let $C$ be Baire space, $\omega^\omega$, with its standard metric?

This is trivial when $Y$ is finite. It also always works when $X$ is compact. An extremely crass way to prove this is to let $g: C \rightarrow H(X)$ be a continuous surjection onto the hyperspace of $X$. Since this function is continuous, and therefore uniformly continuous, with regards to the Hausdorff metric on $H(X)$, the function $f(x,y,c)= g(c)$ is uniformly continuous on $X\times Y \times C$. For any given $y$ we can just choose a $c$ such that $g(c) = F_y$.

It's fairly easy to find examples--even when $X$ and $Y$ are both compact--that show that something like $C$ is necessary, i.e. if we take $C$ to be a one point space then there does not always exist such an $f$. For example if we take $X=Y=[0,1]$ and let $F = \{ (0,0),(1,1) \}$ and $G = \{(x,1-x):x \in [0,1]\}$, then there can be no such $f$ by topological considerations.