Q1 Prove that every simple subgroup of $S_4$ is abelian.
Q2 Using the above result, show that if $G$ is a nonabelian simple group then every proper subgroup of $G$ has index at least $5$.
$|S_4| = 4\cdot3\cdot2 = 2^3 \cdot 3$.
One can exhibit all possible orders of subgroups of $S_4$ and eliminate subgroups according to the requirement that the subgroup is simple.
For example, any subgroup of order $2\cdot 3$ is not simple so needs not to be considered.
Then it might be possible to show that all remaining subgroups are abelian using results like a group of order $p^2$, where $p$ is a prime, is abelian.
I think Cayley's Theorem is involved and I should probably consider the quotient of $G$ on a proper subgroup of $G$. But since $G$ is simple, the quotient is not a group. And I don't really know where to go from here.
For Q1, I'm not sure that the above approach would work and even if it does, I feel like it is too cumbersome and there might be a more principled and more concise way.
Any help would be greatly appreciated.