I have a version of natural irrationalities theorem that states:
Let $F/K$ be a extension of fields. Let's denote by $Gal(f/K)$ the Galois group of the extension $K(\alpha_1,\ldots,\alpha_n)$ for $\alpha_i$ all the roots of the separable polynomial $f \in K[X]$. Similarly for $Gal(f/F)$. Then there is a group monomorphism $Gal(f/F) \to Gal(f/K)$.
From this I should deduce the following corollary:
If Gal(f/K) is simple then for each extension F/K, Gal(f/F) = Gal(f/K) or is trivial.
How can I prove that $Gal(f/F)$ needs to be normal in $Gal(f/K)$?
This isn't true. Indeed, by the fundamental theorem of Galois theory, if $H\subseteq Gal(f/K)$ is any subgroup, then the fixed field $F=K(\alpha_1,\dots,\alpha_n)^H$ satisfies $Gal(f/F)=H$. So to get the result you ask for, you would need $Gal(f/K)$ to have no nontrivial proper subgroups, not just on nontrivial proper normal subgroups (this is equivalent to $Gal(f/K)$ being abelian as well as simple).