Let $G$ be this group and $H$ be any subgroup of index 4.
$G$ acts on the set of left cosets of $H$ in $G$, which is a homomorphism $\varphi: G\to Aut(G/H) = S_4$. It is easy to see that $\ker \varphi\subset H$. We obtain that $$|G/\ker\varphi| = [G:\ker\varphi] = [G:H][H:\ker\varphi] = 4[H:\ker\varphi].$$ But $G/\ker\varphi \hookrightarrow S_4$, thus $|G/\ker\varphi|$ divides $4! = 24$. Therefore $G/\ker \varphi$ can have order $4,8, 12, 24$.
$G/\ker \varphi$ cannot have order $8$, since by Sylow first theorem we have a normal subgroup of order 4, which is index 2. This subgroup lifted to $G$ will be a subgroup of index 2. $G/\ker \varphi$ cannot have order 12, since the Sylow subgroup of 2 is index 3 and can be lifted to a subgroup of index 3 in $G$. $G/\ker \varphi$ cannot have order 24 for similar reason, or one can argument $S_4$ has $A_4$ sitting in it.
Therefore $G/\ker \varphi$ has order 4, meaning $H = \ker \varphi\lhd G$. Please correct me if I am wrong.