Fix any point $p$ in $Y^{reg}$, choose a convex embedded ball $B(p,r)$ in $Y^{reg}$. We claim $B(p(t),r)\cap\Sigma=\emptyset$ for all $t$. For otherwise there is a $T>0$ such that $B(p(t),r)\cap\Sigma=\emptyset$ for $t\in[0,T]$ but $\partial B(p(T),r)\cap\Sigma$ is non-empty. Choose a point $q$ in this intersection. Let $\gamma:[0,1]$ be the radial geodesic connecting $p(T)$ and $q$, and let $p_{i}=\gamma(1-2^{-i})$. Then $B_{i}=B(p_{i},2^{-i}r)\subset B(p(T),r)$, and $d(p_{i},q)=2^{-i}$. Consider the pointed sequence $(Y,2^{i}d_{Y},q)$. By assumption we know as $i$ tends to infinity by passing to a subsequence this converges to a tangent cone $Y_{q}=\mathbb{R}\times\mathbb{C}^{2}/\Gamma$. Then the rescaled balls $2^{i}B_{i}$ converge to a ball $B(p_{\infty},r)$ in $Y_{q}$ and $d(p_{\infty},0)=r$. But $B_{i}$ is isometric to a ball in $B(p,r)$ so have uniformly bounded geometry and thus $2^{i}B_{i}$ converges to a flat ball $B_{\infty}$. Moreover by Lemma 5.3 the distance between any two points in $B_{\infty}$ is realized by the length of a geodesic within $B_{\infty}$. Clearly this can not happen on $Y_{q}$.

By the claim the isometric action $\exp(t\xi)$ is well defined on $Y^{reg}$ for all $t$. Then we can extend the action to an isometric action on $Y$: given $p\in\Sigma$ we pick a Cauchy sequence $p_{i}\in Y^{reg}$ converging to $q$; for any $t$, $p_{i}(t)=\exp(t\xi)$ is also a Cauchy sequence in $Y^{reg}$, so there is a unique limit $p(t)$. We define $exp(t\xi).p=p(t)$. Clearly $\exp(t\xi)$ is distance preserving. Moreover $\exp(t\xi)$ preserves both $Y^{reg}$ and $\Sigma$. ∎