neighbourhood open basis for a connected set in $\mathbb C$

Question

Given $X$ a connected and locally connected space, let $S\subseteq X$ a connected set. Given any open $U$ containing $S$, is it true that there exists an open and connected $V$ such that $S\subseteq V \subseteq U$?

In case the answer is no, is it true if $X = \mathbb R^n$?

Obviously, thanks to the locally connected property, this is true if $S$ is a singleton. Moreover, since $S$ is connected, any couple of disjoint open sets disconnecting $U$ cannot split $S$, so in a sense we can refine $U$ several times, but if we do it infinite times then the resulting $V$ may not be open anymore.
My guess is that if we take for any point $x$ in $S$ an open connected neighbourhood $B_x\subset U$ and then we perform their union, then we have the desired $V$, but I don't know if this process preserves the connectedness.

Answer 1

Your guess is correct, your process preserves connectedness. This can be proved using the fact that for a family of connected subsets with nonempty intersection, the union is connected. Using this lemma twice, each $C_x:=S\cup B_x$ is connected and then, $\cup_{x\in S}C_x$ is connected.