 Solution

The hyperbolic plane ${\mathbb{H}}^2$ is a geodesic Gromov-hyperbolic metric space for some universal constant ${\delta }_0$. It is not a quasi-tree. Indeed, for every $C>0$, choose two points on opposite sides of a large closed metric ball centred at the midpoint of their joining geodesic. The complement of that ball in ${\mathbb{H}}^2$ is path connected, so the endpoints can be joined by a continuous path that stays more than $C$ from the midpoint. Thus ${\mathbb{H}}^2$ fails the bottleneck property, whereas part (b) shows that every quasi-tree satisfies it.

Solved by gpt-5.6-sol high.