Let $C\subseteq\mathbb R^2$ be a closed set with the property that for every $x\in\mathbb R$, there exists at least one $y\in\mathbb R$ such that $(x,y)\in C$.
Does there exist a function $f:\mathbb R\to\mathbb R$ of Baire class 1 such that $(x,f(x))\in C$ for every $x\in\mathbb R$?
EDIT #1: I guess the “Close” vote is because I hadn’t provided more details or outlined an attempt. That is because my hunch was originally split 50–50 between whether the conjecture is true or not. After thinking about it more, I now slightly lean towards thinking the claim is true, at least when the projection of $C$ onto the $y$-axis is compact (so that any counterexample doesn’t hinge on unbounded things).
One particular counterexample I tried constructing was based on the closed topologist’s sine curve. It turns out that that doesn’t work as a counterexample. In particular, let $C$ be the graph of the curve and define $f(x)\equiv\sin(1/x)$ for each $x\in(0,1]$ and let $f(0)\equiv y\in[-1,1]$ be arbitrary. One can construct a sequence $(f_m)_{m\in\mathbb N}$ of continuous functions that converges pointwise to $f$ as follows. Define, for each $m\in\mathbb N$, $f_m(0)\equiv y$, and let $f_m(x)$ coincide with $f(x)$ for $x\in[1/m,1]$. Then, draw a linear interpolation between $(0,y)$ and $(1/m,f(1/m))$.
EDIT #2: Intuitively, the question is (roughly):
Is it true that every closed (or at least compact) set in the plane is arbitrarily close to the graph of some continuous function?