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

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.

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.