Fix any $q\in\Sigma$. Since it is not a fixed point of $\psi$, we can choose a neighborhood $B_{r}(q)$ such that any path-connected component of the intersection of an orbit of $\psi$ with $\overline{B_{r}(q)}$ is compact. Let $O_{q}$ be one path-connected component of $\psi(q)$ in $B_{r}(q)$. We claim for $s>0$ sufficiently small, $\Sigma\cap B_{s}(q)=O_{q}\cap B_{s}(q)$. If not, then there is a sequence $p_{i}\in(B_{r}(q)\setminus O_{q})\cap\Sigma$ converging to $q$. We can choose $q_{i}$ on the path-connected component of the orbit of $p_{i}$ in $\overline{B_{r}(q)}$ which has least distance to $q$. Then $s_{i}=d(q,q_{i})>0$. For $i$ sufficiently large we have $d(q_{i},O_{q})=s_{i}$. Now consider the rescaled pointed sequence $(B_{r}(q),s_{i}^{-1}d_{Y},q)$. As $i\rightarrow\infty$, by passing to a subsequence, this converges to $\mathbb{R}\times\mathbb{C}^{2}/\Gamma$. Moreover, $O_{q}$ converges to $\mathbb{R}\times\{0\}$, and $q_{i}$ converges to $q_{\infty}$ which has distance $1$ to $\mathbb{R}\times\{0\}$. But $q_{i}$ is singular for all $i$, so $q_{\infty}$ is also singular. Contradiction. Then the Proposition follows from the claim and an obvious compactness argument. ∎