$C_2$ are opposite, then $\psi$ does not extend to a $3$-coloring of $G$. Hence, we can assume that the winding numbers of $\psi$ on $C_1$ and $C_2$ are the same. For $i\in\{1,2\}$, let $C_i=v_{i,1}v_{i,2}v_{i,3}$, with the labels chosen so that $\psi(v_{i,j})=j$ for $j\in\{1,2,3\}$. For $j\in\{1,2,3\}$, let $S_{i,j}$ denote the set of vertices of $G$ adjacent to $v_{i,j}$ that do not belong to $V(C_i)$. Since $G$ is tame and the distance between $C_1$ and $C_2$ is at least three, these sets are pairwise disjoint. Let $G'$ be the graph obtained from $G$ by, for $i\in\{1,2\}$ and $j\in\{1,2,3\}$, identifying all vertices in $S_{i,j}$ to a single vertex $x_{i,j}$, and suppressing the resulting faces of length two. Note that $K_i=x_{i,1}x_{i,2}x_{i,3}$ is a non-contractible triangle in $G'$. If $\psi$ extends to a $3$-coloring of $G'$, then it also extends to a $3$-coloring of $G$, obtained by giving each vertex in $S_{i,j}$ the color of $x_{i,j}$.

Let $G_0$ be the subgraph of $G'$ drawn between $K_1$ and $K_2$. By Theorem \[thm-onetri\], there exists a $3$-coloring $\varphi_1$ of $G_0$ such that $\varphi_1(x_{1,j})=j\bmod 3+1$ for $j\in\{1,2,3\}$. By permuting the colors in $\varphi_1$, we obtain a $3$-coloring $\varphi_2$ of $G_0$ such that $\varphi_2(x_{1,j})=(j+1)\bmod 3+1$ for $j\in\{1,2,3\}$ and $\varphi_1(v)\neq\varphi_2(v)$ 