In a locally connected space $X$, one can show that the connected components of any open subspace $U\subseteq X$ are all open in $X$ (cf. Theorem 25.3 in Munkres' Topology 2e). Therefore, if $X$ is also compact, then it has finitely many components.
I am interested in figuring out whether we can figure out the number of components of an arbitrary open subset $O$ of the compact locally connected space $X$. My search is motivated by the following fact:
Let $Y$ be a separable locally connected space. Then every open subspace $W \subseteq Y$ has countably many components.
This holds because $W$ itself must also be separable (by its openness), and local connectedness implies each component of $W$ is open in $Y$, hence intersects the countable dense set $Q \subseteq W$ in at least one point. Since the components of $W$ cover $W$ disjointly, there can therefore only be finitely many of them.
So, I would like a theorem similar to this for $X$ and $O$. I know that it's possible to have $X$ connected and yet $O$ has countably infinitely many components: take $X = [0,1]$ and $O = \bigcup_{n=0}^\infty \left(\frac1{2^{n+1}}, \frac1{2^n}\right)$. My question: is it possible to give a pair of $X$ and $O$ such that $O$ has uncountably many components?
At least in the presence of Countable Choice, such an example would have to have $X$ non-metric, because Countable Choice says that every compact metric space is separable, and then we can apply the above fact about separable spaces. Thus, a side question I have is whether there is a compact locally connected metrizable example of $X$ in ZF without choice.
