Let $G$ be a finite non-abelian simple group and $p$ the largest prime divisor of $|G|$.
If $H<G$, then $|G:H|\geq p$.
Suppose that $|G:H|<p$. Then $|G|<p|H|$.
By Lagrange's Theorem, $|G|=k|H|$.
Hence $1\leq k <p$.
To arrive a contradiction, I have to show that $k=1$ so that $H=G$.
But I do not have idea how can $p$ as the largest prime divisor be used in the proof.
Also, since $G$ is simple group, I can obtain that $G$ can be embedded in $S_{|G:H|}$ but seems not very useful here.
As you noted, if $|G:H|=k<p$, by simplicity we can embed $G$ into $\mathbb S_k$. But $G$ has a Sylow $p$-subgroup of order $p^m$ for some $m\geqslant 1$, and $\mathbb S_k$ doesn't have subgroups of order $p^m$ because $p\not\mid k!$.