I assume that you did the exercise above it that asks you to find all the normal subgroups of $S_n$ ($n\geq 5)$. You should get that the only normal subgroups are the trivial, the alternating, and the whole group.
Now to tackle your problem, let $K$ be a proper subgroup of $S_n$ such that $|S_n/K|=k. Then, note that your group $S_n$ acts on the set of left cosets of $K$ via left multiplication. Thus you get the permutation representation given by: $\pi:S_n\longrightarrow S_k$Note that $\ker(\pi)\leq K$, and recall that kernel of homomorphisms are normal, so you have that $\ker(\pi)$ is a normal subgroup of $S_n$, and thus you have that $\ker(\pi)$ is either trivial, $A_n$ or $S_n$. Note that it cannot be $S_n$ because $K$ has to be proper, and $\ker(\pi)$ cannot be trivial since then you would have an injection from $S_n$ into $S_k$ where $k, a contradiction. The only remaining option is that $\ker(\pi)=A_n$.