the sphere $S(x,s)$ satisfies $\mathcal{H}^1(S(x,s)) < \infty$ and has the property that $\eta \notin \Gamma_0$ for every curve $\eta$ with $|\eta| \subset S(x,s) \cap Q$. Fix such an $s \in (r,2r)$.

Then $B(x,s) \cap Q$ consists of countably many relatively open components $V_j$. By Lemma \[lemm:separating\_continuum\], for such a component $V_j$ there is a simple curve $\gamma_j$ with $|\gamma_j| \subset S(x,s)$ that separates $Q$ into the relative components $U_j$ and $Q \setminus \overline{U}_j$, where $V_j \subset U_j$. Observe that either $\gamma_j$ is a closed curve, or the two endpoints of $\gamma_j$ are contained in $\partial Q$.

Since $B(x,r) \cap Q \subset \bigcup_j U_j$, we have $$\mathcal{H}^1(u(B(x,r) \cap Q)) \leq \sum_j \operatorname{diam}u(U_j).$$ By the maximum principle Lemma \[lemm:maximum\_principle\], $$\operatorname{diam}u(U_j) \leq \sup_{y,z \in \partial_* U_j} |u(y) - u(z)|.$$ By our assumption that $r \leq \min\{ \operatorname{diam}\zeta_1, \operatorname{diam}\zeta_3\}/4$, it follows that if $\zeta_1 \cap \partial_* U_j \neq \emptyset$, then there exists a point $z_1 \in |\gamma_j| \cap \zeta_1$. Indeed, if $y \in \zeta_1 \cap \partial_* U_j$, then $d(y,x) \leq 2r$. But by assumption, there exists $z \in \zeta_1$ such that $d(y,z) > 4r$. The triangle inequality gives $d(z,x) > 2r$, and in particular $z \notin \overline{U}_j$. Since $\gamma_j$ separates $Q$, we conclude there 