In what follows $C_n$ is the cyclic group of order $n$ (also denoted as $\mathbb {Z} _n$ or $\mathbb {Z} /n\mathbb {Z} $), $D_{2n} $ denotes the dihedral group of order $2n$ (with $D_4$ also known as Klein four group). $S_n$ denotes the group of all permutations on $n$ elements and $A_n$ denotes the group of all even permutations on $n$ elements.
We are given that $K$ is a field extension of $\mathbb {Q} $ of degree $4$. Hence we can assume that $K=\mathbb {Q} (a) $ where $a$ is a root of an irreducible polynomial $f(x) \in \mathbb {Q} [x] $ of degree $4$.
And now we can deal with various cases depending on number of roots of $f$ contained in $K=\mathbb {Q} (a) $.
If $K$ contains all the roots of $f(x) $ then $K$ is a Galois extension of $\mathbb {Q}$ and we have the Galois group $\text{Gal} (K/\mathbb {Q} ) $ as either $C_4$ (eg $f(x) =(x^5-1)/(x-1)$) or as $D_4$ (eg $f(x) = x^4-10x^2+1$). In the first case we have just one subgroup namely $C_2$. And $D_4$ has three subgroups each of which looks like $C_2$.
Thus we have either one subfield or three subfields between $\mathbb {Q} $ and $K $.
Next we deal with the case when $K$ contains two roots of $f(x) $. Then the remaining two roots are root of a quadratic polynomial $g(x) \in K[x] $ and hence we can adjoin one of them to $K$ to get the splitting field $L$ of $f(x) $ with $[L:K] =2$.
Next $[L:\mathbb{Q}] =8$ and we know that $\text{Gal} (L/\mathbb {Q}) $ is of order $8$ and contains a transposition (which permutes two roots of $g(x) $) and further the group acts transitively on roots of $f(x) $ (as $f(x) $ is irreducible). Thus the Galois group of $f(x) $ is $D_8$.
The subgroup corresponding to extension $K\subset L$ is $\text{Gal} (L/K) $ and it contains a single transposition apart from identity. A subfield $F$ of $K$ corresponds to a subgroup of $\text{Gal} (L/\mathbb {Q}) \simeq D_8$ containing the subgroup $\text{Gal} (L/K) $. There is exactly one such subgroup and it is isomorphic to Klein-4 group. And thus we have just one subfield $F$. A simple example of this case is the polynomial $f(x) =x^4-2$ (the polynomial $x^4+2$ also works and has same splitting field).
Finally we consider the most complicated case where $f(x) $ has only one root in $K$. The remaining three roots are the roots of a cubic $g(x) \in K[x] $ and $g(x) $ is irreducible (none of its roots lie in $K$). Adjoining one root of $g(x) $ to $K$ we get another field extension $L$ with $[L:K] =3$. It may be possible that $L$ contains all roots of $g(x) $ in which case $L$ is the splitting field of $f(x) $ with $[L:\mathbb{Q}] =12$ and the Galois group of $f(x) $ is isomorphic to $A_4$.
Now $\text{Gal} (L/K) $ corresponds to a subgroup of $A_4$ of order $3$ and there is just one such subgroup which is isomorphic to $A_3$ (equivalently to $C_3$).
A subfield $F$ of $K$ corresponds to a proper subgroup of $\text{Gal} (L/\mathbb{Q}) $ which contains $\text{Gal} (L/K) $ properly. Thus its order must be $6$. But $A_4$ has no subgroup of order $6$ and hence such a field like $F$ does not exist. An example of such a case is $f(x) = x^4+8x+12$ (taken from notes by Keith Conrad).
We still have to deal with the case when $L$ does not contain all the roots of $g(x) $. Then $L$ contains just one root of $g(x) $ and remaining two roots are the roots of a quadratic polynomial $h(x) \in L[x] $. Adjoining one root of $h(x) $ to $L$ we get the splitting field $M $ of $f(x) $ with $[M:L] =2$ and $[M:\mathbb{Q}] =24$ so that the Galois group of $f(x)$ is isomorphic to $S_4$.
The field $K$ corresponds to a subgroup of order $6$ of $S_4$ and there is only one such subgroup which is isomorphic to $S_3$.
A subfield $F$ of $K$ corresponds to a proper subgroup of $S_4$ which contains $S_3$ properly. Hence its order must be $12$. The only subgroup of order $12$ is $A_4$ and it consists of even permutations only and hence it can't contain $S_3$. Thus we don't have the desired subgroup and hence no such desired subfield $F$ exists. An example for this case is given by $f(x) =x^4+x+1$ (this is also mentioned in comments to question, Keith Conrad mentions $x^4-x-1$ as another example in his notes).
