SOLVED - check out the comments or directly this answer.
I was wondering if the boundary of an unbounded component $C$ of the complement of a bounded connected open set $U$ must be connected (in $\mathbb R^n, n\geq 2$).
As the complement of $C$ (i.e., the sum of $U$ and the bounded components of its complement) is connected, the answer is positive if the following claim is true:
A connected open set $A$ in $\mathbb R^n$ $(n\geq 1)$ has disconnected complement if and only if it has disconnected boundary.
There is a proof here, but one crucial step seems false (see below). I'm looking for a correct proof (or a counterexample, but I hope it is true).
I had two doubts about the proof mentioned - one of them is the above probably false step and the other is already solved:
Why can we join $x$ and $y$ through the complement of $A$? We would need something like connectedness of interior of this complement, wouldn't we? UPDATE: YES, see below.
Why this homotopy cannot exist? UPDATE2: solved, see below.
Update: The statement about joining $x$ and $y$ seems false: even if $A$ and its complement are connected and the boundary is connected with $x,y \in \partial A \cap \partial\mathrm{int} (\mathbb R^n \setminus A)$, it is not true, that we can join $x,y$ by a curve in $X\setminus A$. It is enough to consider a Warsaw sinusoid in the plane with closed balls attached to its different path-components, and take $x$ and $y$ at the boundaries of these balls.
Note that we can improve this example so that any point in the boundary of $A$ is in the boundary of interior of its complement - so the assumption about regular openness of $A$ proposed in the original question doesn't help.
Update2: The argument for non-existence of the homotopy trivialising the loop can be derived from the tricky function proposed by Dejan Govc in another answer to the original question (in the part entitled "Added"). The loop composed with this function is a nontrivial loop in $S^1$, so the original loop also is nontrivial, a contradiction.
