Let $E$ be a finite extension of $F$ and assume that $E/F$ is separable. Since $E/F$ is separable, there is a finite Galois extension $E'$ over $F$, containing $E$: $F \subseteq E \subseteq E'$. Show that
$E' \cap \bar{F} = F$
$x \in F \implies x \in E'$ (because $F \subseteq E'$) $\land \ x\in \bar{F}$ (because $F \subseteq \bar{F}$) $\implies x \in E' \cap \bar{F}$
I would like to ask for a tip to show other direction.
Let's see: $\;E=\Bbb Q(\sqrt[4]2)\;$ is a finite extension of $\;F=\Bbb Q\;$, and of course $\;E/F\;$ is separable. $\;E'=\Bbb Q(\sqrt[4]2\,,\,i)\;$ certainly contains $\;E\;$ and $\;E'/F\;$ is Galois .
But do you think $\;E'\cap \overline F=\Bbb Q(\sqrt[4]2,\,i)\cap\overline{\Bbb Q}=F=\Bbb Q\;$ ? In fact, it must be $\;E'\cap\overline F=E'\;$ in any case with the given data, since $\;E'\;$ algebraic over $\;F\;$ ...!