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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.

A subgroup is quasiconvex when it is a quasiconvex subset of a Cayley graph. In a hyperbolic group, a finitely generated subgroup is quasiconvex exactly when it is quasi-isometrically embedded.