A subset of $X$ an indecomposable continuum $Y$ is called a composant transversal if $X$ has exactly one point from each composant of $Y$.
So a continuum has a composant transversal precisely when there exists a choice function on its set of composants.
Solecki proved that no indecomposable continuum has a Borel composant transversal.
Solecki, Sławomir, The space of composants of an indecomposable continuum, Adv. Math. 166, No. 2, 149-192 (2002). ZBL1014.54021.
But what if we ask something a little different:
Main Question. Let $X$ be a dense Polish subspace of an indecomposable continuum $Y$. Must some composant of $Y$ contain at least two points of $X$?
If "Borel" replaces "Polish", then I have an example showing the answer is no. But I suspect the answer is yes. For instance, it is not difficult to show that a dense $G_\delta$-subset of $C\times[0,1]$ must contain uncountably many points from each of uncountably many arcs $\{c\}\times[0,1]$.
Ideas: If the union of all composants touching $X$ is Polish, then I think Solecki's results show the answer is yes. But the most I can show is that $X$ is a countable union of $G_\delta$-sets. Another idea is to suppose $X$ is a Polish "partial transversal", and then show $X$ can be extended to a Borel composant transversal, reaching a contradiction.
On another note, by Solecki's result the existence of composant transversals seems to require something close to the full-blown axiom of choice, or at least AC$(\mathbb R)$.
Other questions.
Are composant transversals non-measurable?
Does something like ZF$+$AD$+$DC imply there is no composant transversal?