Let $U \subset \mathbb R^2$ be an open, regular (meaning $U$ is the interior of its closure), bounded and simply connected set. Let $\gamma : [0,1] \rightarrow \partial U$ be a simple curve such that $\gamma(0) \neq \gamma(1)$.
Is $\partial U \setminus \gamma((0,1))$ still connected ?
I think this should be true. Assume for a contradiction that this is not the case. Let $K$ be the connected component $\{\gamma(0)\} \subset K \subsetneq \partial U \setminus \gamma((0,1))$, where we remark that $K$ and $\partial U \setminus \gamma((0,1))$ are compact. By Zoretti's Theorem, for any $\varepsilon > 0$, we can find a simple closed curve $J : [0,1] \rightarrow \mathbb R^2$ for which $$ K \subset \mathrm{int}(J), \quad J \cap \left( \partial U \setminus \gamma((0,1)) \right) = \emptyset, \quad \mathrm{dist}(K, j) \leq \varepsilon \quad \forall j\in J $$
Note that $J$ cannot be contained in $U$ (otherwise, $K \subset \mathrm{int}(J) \subset U$ which contradicts the regularity of $U$) and $J$ can only cross $\partial U$ through $\gamma((0,1))$.
I think we should be able to deduce from this that $\gamma(0)$ would actually lie on the exterior of $J$ but I am not able to complete my proof.
Any help is welcomed.
EDIT : This is related to the notion of cross-cut. A cross-cut $\phi$ leads to a decomposition of $U$ into two simply connected domains, whose boundaries are then connected and contain $\phi$. What would happen to these boundaries if one removed $\phi$ ?
EDIT 2 : My above argument is incorrect because $\gamma((0,1))$ is not necessarily open in $\partial U$ topology and this is what leads to the counter-example below.
