I am trying to prove the following assertion: Let $B$ be a simply-connected set, and let $B' \subsetneq B$ be a proper open connected subset. Then, there exists a point $x \in \partial B'$ of the boundary such that there is a basis of neighborhoods of $x$ in $B$ whose intersections with $B'$ are connected.


This seems like it should be true. For example, in the one-dimensional case, a proper open subinterval of an interval always has an endpoint satisfying this property, whereas for a circle, one can take the complement of a point $x \in S^1$, and the boundary point $x$ does not have the property in question.


I am happy if one takes the special case of $B$ a manifold, or even a product of intervals.



Boundaries of subsets of simply-connected domains
It seems if you take $B=\mathbb{R}^2$ and $B'$ the complement of the closure of $\Big\{\big(x,\sin\big(\frac{1}{x}\big)\big);x\in(0,\infty)\Big\}$ this is a counterexample. (Added bonus: $B'$ is also simply connected)


