 Solution

Let $p$ be the midpoint of a geodesic $[y_1,y_2]\subseteq Y$ and put $L=d(y_1,y_2)$. Apply the geodesic quadrilateral in a hyperbolic metric space bound to the quadrilateral with consecutive vertices $y_1,z_1,z_2,y_2$. The point $p$ is within $2\delta$ of one of the other three sides. It cannot be within $2\delta$ of $[z_1,z_2]\subseteq Z$, because every point of $Y$ has distance greater than $2\delta$ from every point of $Z$.

By symmetry there is therefore a point $q\in [y_1,z_1]$ with $d(p,q)\le 2\delta$. The triangle inequality givesand henceSince $p\in Y$ and $y_1$ is a closest point of $Y$ to $z_1$, we also have $d(z_1,y_1)\le d(z_1,p)$. Therefore $L\le 8\delta$, as required.