It is required to prove that $S_n$ has no subgroup of index $t$ for $2 < t < n$.
The main difficulty is on the case where $n=4$ and $t=3$.
What has been done:
By contradiction, suppose there is $H\le S_n$ such that $[S_n: H] = t$, for $2 < t < n$. Since $S_n$ is finite, by Lagrange Theorem $|H| = \frac{n!}{t}$.
Now, by Cayley, every time $H\le G$ such that $[G:H]=n$, there is an homomorphism $\phi: G \to S_n$ such that $\ker \phi \le H$.
Thus, there is an homomorphism $\phi: S_n \to S_t$ such that $\ker \phi \le H$. For $n\neq 4$ it happens that $A_n$ is the unique non trivial normal subgroup of $S_n$ and as $\ker \phi \trianglelefteq S_n$, $\ker \phi = \{1\}$ or $\ker \phi = A_n$.
If $\ker \phi = \{1\}$, $\phi$ is injective and $|S_n| \le |S_t|$, which is a contradiction since $t<n$
So, let $\ker(\phi) = A_n$, where $[S_n: A_n] = \frac{n!}{2}$. Since $S_n$ is finite, by Lagrange Theorem, given $A_n \le H \le S_n$, $$\frac{n!}{2} = [S_n: A_n] = [S_n: H] [H: A_n] = t [H: A_n] \longrightarrow [H: A_n] = \frac{n!}{2t}$$
Once again, by Lagrange Theorem, since $A_n \le H$, $$|H| = |A_n| [H: A_n] = \frac{n!}{2} \frac{n!}{2t}$$
Then $$\frac{n!}{t} = |H| = \frac{(n!)^2}{4t} \longrightarrow n! = 4$$ which is a contradiction.
So, it's missing the case where $n = 4$ and $t = 3$. Thus, let $H\le S_4$ such that $|H| = 8$ and $[S_4: H] = 3$.
Any valuable help is appreciated.
Related:
Show that $S_n$ has no subgroup of index $t$ for $2 < t < n$
EDIT 1:
By Cayley, there is $\phi: S_4 \to S_3$ such that $\ker \phi \le H$. Thus, $|\ker \phi| \in \{1, 2, 4, 8\}$.
If $|\ker \phi| = 1$, $\ker \phi = \{1\}$. Therefore, $\phi$ is injective and $|S_4| \le |S_3|$, which is a contradiction.
Since $\vert S_4 \vert = 24=2^3 \cdot 3, S_4$ has a $2$-Sylow subgroup of order $8$ and index $3$. I'll leave finding one as an exercise for you.