= Solution
Take $G=\mathbb Z^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and let
$$
H=\langle(1,1)\rangle.
$$
Intrinsic 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 subgroup>[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.
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