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Let $f(x) \in \mathbb{Q}[x]$ be an irreducible quintic polynomial with exactly 2 nonreal zeros. If $E$ is the splitting field of $f(x)$, then there exists a Galois extension $K$ of $\mathbb{Q}$ with $K \subset E$ such that:

a-) $[K:\mathbb{Q}]=2$

b-) For any field $F \neq E$,$F\neq K$ with $\mathbb{Q} \subset K \subset F \subset E$, $F$ is not a Galois extension of $\mathbb{Q}$

I know that the Galois group of $f(x)$ is $S_5$ and therefore $f(x)$ is not solvable by radicals, but I do not know how to construct this extension and any help is appreciated. Thanks in advance.

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The fundamental theorem of Galois Theory establishes a bijection $H\mapsto E^H$ between subgroups of $Gal(E/\mathbb{Q})$ and subfields of $E$ containing $\mathbb{Q}$. Moreover, the properties of this bijections give you that if $K=E^H$, then $K$ is a Galois extension if and only if $H$ is normal, $[K:\mathbb{Q}]=2$ if and only if $[Gal(E/\mathbb{Q}):H]=2$ and condition b) is equivalent to "for all $H'$ subgroup of $Gal(E/\mathbb{Q})$ with $Gal(E/\mathbb{Q})\supset H \supset H' \supset \{1\} $ (where all inclusions are proper), $H'$ is not normal".

Thus the question is equivalent to searching for a normal subgroup $H$ of index $2$ which is also simple. $H=A_5$ fits all these; thus $K=E^{A_5}$ is the solution.

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