Suppose that for every $x\in[0,1]$, $A_x$ is a dense subset of $\mathbb{R}$. Assume the axiom of choice holds.
Is there a way to construct a continuous function $f$ over $[0,1]$ so that for all $x$, $f(x)\in A_x$?
My first instinct was to pick any $f(0)$ and $f(1)$, and then define intermediate values by induction on $n$ so that $\left|f\left((2k+1)2^{-n}\right) - \frac{f\left(k2^{-n+1}\right)+f\left((k+1)2^{-n+1}\right)}{2}\right| < (2+\epsilon)^{-n}$ $f\left((2k+1)2^{-n}\right) \in A_{(2k+1)2^{-n}}$ which can then be uniquely extended by continuity to $[0,1]$, but unfortunately this extended function does not satisfy $f(x)\in A_x$ in the general case.