Here's a ZF proof that if $S$ is a chain of sets with $\bigcup S = \mathbb{R},$ then there is $X \in S$ which contains a countable set dense in some nonempty open set.
If there is $X \in S$ such that $\mathbb{Q} \subset X,$ we are done. Assume this is not the case. Let $\langle q_n \rangle$ be an enumeration of $\mathbb{Q}.$ Let $$X_n = \bigcap \{X \in S: \forall i<n (q_i \in X)\}.$$ Let $Y_n = X_{n+1} \setminus X_n.$ Then $\langle Y_n \rangle$ is a countable partition of $\mathbb{R}.$ I show in the first part of my answer here that some $Y_n$ contains a countable dense subset of an interval. Then any $X \in S$ containing $\{q_0, \ldots, q_n\}$ is as desired. $\square$
Generalizing along a different axis, if $S$ is a chain of subsets of $\mathbb{R}$ each discrete under the subspace topology, then $\bigcup S$ is countable. Let $\langle U_n \rangle$ enumerate a topological basis. Then we can construct a surjective partial map $f: \omega \rightharpoonup \bigcup S$ by $f(n) = x$ if there is $X \in S$ such that $U_n \cap X = \{x\}.$
Both of the above readily generalize to an arbitrary Polish space.