= 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.
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