Conditions for a intersection of connected sets to be connected.

This is the original statement I want to prove:

Conjecture: Let $A$ and $B$ be open connected sets in $\mathbb R^2$ with the usual metric such that $A \cap B \neq \emptyset$. Then $ \partial A \cap \partial B = \emptyset \Longrightarrow$ $ A \cap B $ is connected.

It is a simple statement, but I cannot find a proof neither a counter example. Any ideas are welcome!

Update 0:

Some comments about this being a general topological result made me realize that an example of space where this doesn't hold is important as motivation.

Here is a simple example of such space:

Consider the circle $S^1 = \mathbb R / \mathbb Z$ and the intervals $A = (\frac{2}{8}, \frac{7}{8})/ \sim$ and $B = (\frac{-3}{8}, \frac{3}{8})/ \sim$ ($\sim$ is the equivalence relation defining the circle). Note that they satisfy the conditions but their intersection is not connected: $A \cap B =( (\frac{2}{8}, \frac{7}{8}) \cup (\frac{5}{8}, \frac{7}{8}))/ \sim$. This is illustrated below.

Update 1:

I came with a weaker version of this statement that is sufficient for the application I need.

Weak Conjecture: Let $A,B$ be open connected sets in $\mathbb R^2$ such that $A \cap B \neq \emptyset$ and each connected component of $\partial A$ and $\partial B$ is the image of a proper embedding from $\mathbb R$ to $\mathbb R^2$. Then $ \partial A \cap \partial B = \emptyset \Longrightarrow$ $ A \cap B $ is connected.

$\ \ \ $Proof:

$\quad$ First, we will refer to images of proper embeddings from $\mathbb R$ to $\mathbb R^2$ as lines. Note that lines splits the plane into two connected components. In the case that a line $\phi$ is a connected component of the boundary of some connected set $U$, we can define $D(\phi)$ as the connected component of $\mathbb R^2 \setminus \phi$ which contains $U$. The other c.c. is denoted $E(\phi)$. If we denote $\Phi_U$ the collection of all c.c. of $\partial U$, we can write

$$ U = \bigcap_{\phi_i \in \Phi_U} D(\phi_i) \qquad \ \ \ \ \ U^c = \bigsqcup_{\phi_i \in \Phi_U} E(\phi_i)\;. $$ It is important to note that the union above is disjoint. Otherwise, we would get $i_1 \neq i_2$ such that $\phi_{i_1} \cap \phi_{i_2} \neq \emptyset$, absurd since they are different c.c. components of $\partial U$.

$\quad$ Now we begin the proof. Let $C : = A \cap B$ and suppose $C$ has at least two nonempty connected components $C_1$ and $C_2$. Note that $ C_2 \subset C_1^c = \bigsqcup_{\phi_{i} \in \Phi_{C_1}} E(\phi_i)$, therefore $\exists ! \ i_n$ s.t. $C_2 \in E(\phi_{i_n})$. Since $\partial A \cap \partial B = \emptyset $, we can suppose without loss of generality that $\phi_{i_n} \subset \partial A$. However, $C_1 \subset A$ and $C_2 \subset A$ are contained on different c.c. of $\mathbb R^2 \setminus \phi_{i_n}$, which would imply that $A$ is not connected, absurd.

$\qquad $Q.E.D.

The main question remains open for discussion. The understanding of the topology of $\partial U$ is all that we should need to conclude it.

I'll try to sketch some ideas that might lead to a proof. The following is definitely incomplete, also too long for a comment. Did you consider anything along these lines?

Assume by way of contradiction that the hypotheses hold but $A\cap B$ is not connected. Then take two points $x,y\in A\cap B$ in different connected components. Since connected open sets in $\mathbb{R}^2$ are arcwise connected, we may find a path $P_A:[0,1]\to A$ from $x$ to $y$ inside $A$. Considering that $\partial A\cap \partial B=\emptyset$, we might assume the starts in $A\cap B$, traverses $A\setminus B$ and reaches $A\cap B$ in that order:

(To see this, it should be enough to find a maximal subinterval of $P_A^{-1}(A\smallsetminus B)$, which must exist, otherwise, $B$ would accumulate along the path and we would have a common boundary point.) The point $a$ lies in the boundary of $A\cap B$: more precisely, in $A \cap \partial(A\cap B) \smallsetminus B$.

Similarly we can find a path $P_B:[0,1] \to B$ from $x$ to $y$ (though now I won't be confident that I can have the same property as above). At least, we can be sure that we have some point $b$ in $B \cap \partial(A\cap B) \smallsetminus A$.

Now, let us restrict ourselves to an open ball $V$ containing $x$, $y$, and both paths. It is easy to see, since $A$ and $B$ are open, that every point of $\partial (A\cap B)$ must be on $A\smallsetminus B$ or $B\smallsetminus A$, otherwise it would lie in $\partial A\cap \partial B$. So (using compactness) we can cover $\partial (A\cap B)\cap\overline V$ by finitely many separated balls contained in $A$ or in $B$.

If we focus on the balls containing $a$ and $b$, we have a situation as pictured below:

where red corresponds to $A$ and blue to $B$, and the left ball is completely included in $B$ and the right ball inside $A$. This looks rather strange, since $x$ lies in (the interior of) $A\cap B$, and therefore we should be able to find some point of $\partial (A\cap B)$ in the region between the two paths.