Let $L|K$ be a finite Galois extension and let $E$ and $F$ subextensions of $L|K$. I have to see that $$\text{Gal}(L|EF) = \text{Gal}(L|E) \cap \text{Gal}(L|F).$$
It is clear that $\text{Gal}(L|EF) \subseteq \text{Gal}(L|E) \cap \text{Gal}(L|F)$. What I don't see so clearly is the other inclusion. Could you help me with that, please?
Note: $EF = K(E \cup F)$ is the smallest subextension of $L|K$ which contains $E$ and $F$.
Because $E$ is a finite extension of $K$, there exist elements $e_1, \dots, e_m \in E$, which are algebraic over $K$, such that $E = K(e_1, \dots, e_m)$.
And therefore, any element of $E$ can be written as a $K$-linear combination of monomials of the form $e_1^{n_1}\dots e_m^{n_m}$, where each $n_i$ is an integer less than the degree of $e_i$ over $K$.
Using this observation, perhaps you can convince yourself that an automorphism $\sigma$ of $L$ fixes all elements of $E$ if and only if it fixes the elements $e_1, \dots, e_m$.
Of course, similar comments apply to $F$: there exist elements $f_1, \dots, f_n \in F$ such that $F = K(f_1, \dots, f_n)$, and such that an automorphism $\sigma$ of $L$ fixes all elements of $F$ if and only if it fixes $f_1, \dots, f_n$.
Finally you can try to argue that $EF = K(e_1, \dots, e_m, f_1, \dots, f_n)$. And then go from there.