\cap E \neq \emptyset, \quad (Q \setminus U) \cap E \neq \emptyset, \quad \partial U \cap E = \emptyset.$$ Let $E_1 = U \cap E$ and $E_2 = (Q \setminus U) \cap E$. By passing to a subset if needed, we may assume that $E_1$ and $E_2$ are each contained within a single component of $U$ and $Q \setminus \overline{U}$, respectively. We fix $\varepsilon >0$ such that $\operatorname{dist}(\partial U,E) > \varepsilon$. By Proposition \[prop:coarea\] applied to $h(x)=\operatorname{dist}(\partial U,x)$, there is $0<p<\varepsilon$ such that $\mathcal{H}^1(h^{-1}(p))< \infty$ and every rectifiable curve $\gamma$ for which $|\gamma| \subset h^{-1}(p)$ lies outside the exceptional set $\Gamma_0$. By and our choice of $p$, the sets $E_1$ and $E_2$ belong to different components of $Q \setminus h^{-1}(p)$. Lemma \[lemm:separating\_continuum\] then shows that $h^{-1}(p)$ has a connected subset $F \subset Q$ such that $E_1$ and $E_2$ belong to different components of $Q \setminus F$. Notice that, for every rectifiable curve $\gamma$ with $|\gamma| \subset F$, $u||\gamma|$ is continuous and either $u(x) < s$ or $u(x)>t$ for all $x \in |\gamma|$. We divide the rest of the proof into cases.

\[pring\] Suppose there is an open set $G \subset X$ such that $\partial G \subset F$ and $E_j \subset 