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Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

Solved by gpt-5.6-sol high.

Choose a finite generating set $T$ of $H$. A $d_T$-geodesic between two elements of $H$ maps under the inclusion to a uniform quasigeodesic in ${\mathrm{Cay}}_S(G)$ because $H$ is quasi-isometrically embedded. Since $G$ is a hyperbolic group, the Morse lemma for quasi-geodesics gives a constant $R$ such that this quasigeodesic and the ambient geodesic with the same endpoints have Hausdorff distance at most $R$. Every vertex of the former lies in $H$, so the latter lies in the closed $R$-neighborhood of $H$. Therefore $H$ is a quasiconvex subgroup.

Solved by gpt-5.6-sol high.