Extending a connected open set Assume $\emptyset\neq V\subseteq U\subseteq\mathbb{R}^n$ are open and connected sets so that $U\setminus\overline{V}$ is connected as well. Given any point $x\in U$, is there always a connected open set $W\subseteq U$ so that $\{x\}\cup V\subseteq W$ and $U\setminus\overline{W}$ is connected? In other words, can $V$ be extended to a connected open set containing a given point so that the complement of the closure of the extended set is still connected?
 A: There are two cases to consider:


*

*$n\ge 2$. Observe that in this case, for every open connected set $S\subset R^n$, any $a\in A$ and any sufficiently small $r\ge 0$, the complement of the closed ball 
$$
A\setminus \overline{B(a, r)}
$$
is still connected. Now, if $V$ is dense in $U$ then the only meaningful answer is to take $W=U$ (for any choice of $x$) and then $U\setminus \bar W$ and $W$ are both connected (the first one is empty of course). If $V$ is not dense in $U$, we can do a bit better than this: The subset $U\setminus \{x\}\cup \bar V$ is open and nonempty. Pick any point $a$ in this complementary set and let
$$
W= U \setminus \overline{B(a, r)}
$$
where $r>0$ is sufficiently small. Then 
$$
U\setminus \bar W = B(a, r)
$$
(the open ball) is connected and nonempty and $W$ is also connected by the above remark. 

*$n=1$ (I will leave out the case $n=0$). Then both $U$ and $V$ are intervals an you can take $W$ to be the smallest open interval containing $V$ and $x$.  
A: Note that it is enough to show this only for $x ∈ ∂V ∩ U$. Consider $A := \{x ∈ U: ∃W ⊆ U \text{ connected, } \{x\} ∪ V ⊆ W, U \setminus \overline{W} \text{ connected}\}$ This is obviously open subset of $U$. If it is also closed then it's whole $U$.
A: First, trivial solution is $W:=U$ (since the empty set $U\setminus \bar V$ is connected).
Second, provided you wish to have $U\setminus \bar W$ nonempty, there is still trivial solution: You obviously add an assumption $U\setminus \bar V \neq \emptyset$, since
otherwise the statement is wrong. Now $O:=U\setminus \bar V$ is non-empty open set, necessarily infinite, so we can choose a point $y\in O \setminus \{x\}$.
Now if $r:=\operatorname{dist}(y, V\cup \{x\} \cup (\mathbb R^n \setminus U)) $ and $W=U \setminus \bar B(y,r/2)$,
then $U\setminus \bar W = \bar B(y,r/2)$ is non-empty and connected as required.
May be you want some more properties on $U$, for which purpose I add some proofs, to which you could add more to obtain your properties. May be you think about $W$ being simply connected provided $V$ is simply connected or the like.
I offer two proofs in addition to the above two.
The first  one is short but a bit abstract.
Let $$
A:=\{ x \in U : \exists W_x \subset U, W_x \text{open, connected}, U\setminus \bar W_x \text{connected} \}
$$ 
Then $V\subset A$ since for any $x\in V$, $W_x:=V$ is a good choice. Thus $A$ is non-empty.
$A$ is open in $U$ since if $x\in A \subset U$, $W_x$ is as in the definition of $A$ and $B(x,2r) \subset U$, $r>0$, then for every $y\in B(x,r/3)\cap U$, we let $L_y$ be the shortest segment connecting $y$ to $W_x$. Since $x\in W_x$, the length of $L_y$ is less then $r/3$ and $L_y\subset B(x,2r/3)$. We let $W^*_y$ be the union of $W_x$ and the $r/3$-neighbourhood of $L_x$. Let $W_y$ be the union of $W^*_y$ and every component $U\setminus \bar W^*_y$ that is contained in $B(x,r)$. (I am sorry, this is not very elegant.). Thus $y\in A$ and $B(x,r/3)\cap U\subset A$, showing that $A$ is open in $U$.
$A$ is also closed in $U$: if $y\in U \cap \bar A$, then we choose $r>0$
such that $B(y,2r) \subset U$ and $x\in A$, $\| x-y \| < r/3$. Adding the segment $xy$ and its $r/3$-neighbourhood to $W_x$, together with any trashy components the same way as in the previous paragraph, we obtain a set $W_y$ that shows that $y$ belongs to $A$.
Thus $A$ is closed in $U$.
Since $U$ is connected, and $A$ is non-empty, and both closed in open in $U$, we have $A=U$ which is what was to be proved.
The last proof will be most easily understandable, but I will not put much details, leaving it very sketchy.
The case $V=\emptyset$ is easy (put a small ball around the given point $x$). We assume $V$ is non-empty.
Let $x\in U$, and choose any point $y$ in $V$.
Recall that open connected set $U$ in $\mathbb R^n$ is also path connected, and for every $x, y \in U$, there is path consisting of finite number of segments. Moreover the segments in the paths can be disjoint, with the obvious exception of endpoints of 'consecutive' segments.
Thus let $p$ be such a path connecting $x$ and $y$. Let $z$ be the point on $p$
which is the first point that is in $\bar V$ is we go from $x$ to $y$.
Let $q$ be the part of $p$ that connects $x$ to $z$. $q$ consists of finitely many segments, all of them with the exception of last do not intersect $\bar V$.
To each of the segments we add a small neighbourhood that does not intersect 1. $\bar V$
2. closures of neighbourhoods of other segments, with the exception of the consecutive segments. We also add a neighbourhood of the segment of $q$ that contains $z$, and solve somehow the problem that can be created by $V$ having possibly bad shape near $z$ (see the trashy components in the previous proof.) This way we obtain $W$.
I hope this helps you to understand your own question properly as well as how the proofs might look.
