Does connected open subset intersect another subset exactly when it intersects the subset and its complement? Let $(X, \mathcal{T})$ be a topological space, $U \in \mathcal{T}$ be connected, and $V \subset X$. Does it hold that
$U \cap \partial V(X) \neq \emptyset \iff U \cap V \neq \emptyset \text{ and } U \setminus V \neq \emptyset$?
Here $\partial V(X)$ is the boundary of $V$ in $X$. Similarly $\bullet V(X)$ is the interior of $V$ in $X$, and $\overline{V}(X)$ is the closure of $V$ in $X$.
Background
I can prove that the following are equivalent:

*

*$U \cap \partial V(X) \neq \emptyset$

*$U \cap V \neq \emptyset \text{ and } U \setminus \overline{V}(X) \neq \emptyset$

*$U \cap \bullet V(X) \neq \emptyset \text{ and } U \setminus V \neq \emptyset$
Also, the direction $\implies$ holds even without assuming $U$ is connected, so the question is actually about the $\impliedby$ direction.
In particular, the above shows that the claim holds when $V$ is either open or closed. I suspect there is a counter-example for general $V$, but cannot come up with one.
 A: Yes, what you say is true.
The direction “$\implies$” is true even if $U$ is not connected. Indeed, let $x\in U\cap\partial V$. Since $x\in\partial V$, every neighborhood of $x$ has nonempty intersection with both $V$ and $X\setminus V$. Since $U$ is a neighborhood of $x$, the latter is true for $U$.
The direction “$\Longleftarrow$” is true even if $U$ is not open. Indeed, suppose that $U$ is connected and both $U\cap V$ and $U\setminus V$ are nonempty, but $U\cap\partial V=\emptyset$. Then,
$$U\cap V = U\cap (V\setminus\partial V) = U\cap\mathring{V}\neq\emptyset$$
and also, denoting $W=X\setminus V$
$$U\setminus V = U\cap W = U\cap(W\setminus\partial V) = U\cap(W\setminus\partial W) = U\cap \mathring{W}\neq\emptyset$$
Note that we have arrived at the following situation: $U$ is the disjoint union of $U\cap V=U\cap\mathring{V}$ and $U\setminus V = U\cap\mathring{W}$, which are both open in $U$ (since both $\mathring V$ and $\mathring W$ are open) and nonempty, i.e., $U$ is disconnected: contradiction.
A: For the $\impliedby$ direction:
Suppose $U \cap V \neq \emptyset$ and $U \setminus V \neq \emptyset$, but $\partial V \cap U = \emptyset$. For each $u \in U$, since $u$ is not a boundary point of $V$, there is a neighborhood $B_u \subset U$ of $u$ which is either entirely contained in $V$, or entirely not contained in $V$. If $W = \bigcup_{u \in U \cap V} B_u$, and $W' = \bigcup_{u \in U \setminus V} B_u$, then both are non-empty open sets contained in $U$, and $U = W \cup W'$. So $U$ is disconnected.
