Assume that $K$ is a normal extension of F, and $G(K/F)$ is isomorphic to $S_4$. How many fields E are there for which $F \subset E \subset K$ and $[K : E] = 2$?
The only two subgroups of order 2 of $S_4$ are $\{id, (1,2)\}$ and $\{id, (1,2), (3,4)\}$. But I am not sure how to determine if they are the subgroup of $G(K/F)$. Any help would be appreciated!
Standard result from Galois theory: if $K/F$ is a finite Galois extension, then $K$ is Galois over every intermediate field of $K/F$, and $E \mapsto \textrm{Gal}(K/E)$ is a bijection from the set of such fields onto the set of subgroups of $\textrm{Gal}(K/F)$, under which you have $[K : E] = |\textrm{Gal}(K/E)|$.
Your hypothesis being $\textrm{Gal}(K/E) \cong S_4$, you had one job: count the number of subgroups of $S_4$ whose order is two, i.e. count the elements of order $2$.