\zeta_j$, where either $i \in \{1,3\}$ and $j \in \{2,4\}$, or $j \in \{1,3\}$ and $i \in \{2,4\}$.

Suppose that $x_0 \in \zeta_1$ and $x_1 \in \zeta_3$. Then, since $\gamma \notin \Gamma_0$, the construction of $u$ shows that $u||\gamma|$ takes all values between $0$ and $1$. In particular, $u(x) \in [s,t]$ for some $x \in |\gamma|$. This contradicts the fact that $|\gamma| \cap E = \emptyset$. The argument remains valid if the roles of $x_0$ and $x_1$ are reversed.

Suppose that $x_0 \in \zeta_2$ and $x_1 \in \zeta_4$. Without loss of generality, we may assume that $\Omega_1$ is the component containing $\zeta_1$. It then follows from Lemma \[lemm:maximum\_principle\] that $u(x) \geq s$ for some $x \in |\gamma|$. Moreover, since $u||\gamma|$ is continuous and $|\gamma| \cap E = \emptyset$, it follows that in fact $u(x) >t$ for every $x \in |\gamma|$. Similarly, applying Lemma \[lemm:maximum\_principle\] to $\Omega_2$ shows that $u(x) < s$ for every $x \in |\gamma|$. This is a contradiction. The argument remains valid if the roles of $x_0$ and $x_1$ are reversed.

We conclude that the set $E$ is connected. It remains to show that $E$ intersects both $\zeta_2$ and $\zeta_4$. Suppose towards contradiction that this is not the 