I completely edited my question and with the help of Lubin, I added my proof. Any comments are welcome.
My question was:
Let $F_0 \subset F_1 \subset F_2$ be fields, and suppose $F_2/F_0$ is a Galois group.
Then, Galois group Gal($F_2/F_1$) is a normal subgroup $\iff$ $F_1 /F_0$ is a Galois extension.
How to prove normal $\Rightarrow$ Galois?
This is my revised proof:
First, $F_1=$ $ F_2^{\text{Gal}(F_2/F_1)}$, where $ F_2^{\text{Gal}(F_2/F_1)}$ is ($F_2$ fixed by Gal$(F_2/F_1)$) ,
Since $F_1 \subset F_2^{\text{Gal}(F_2/F_1)}$, and $[F_2 : F_1] = |\text{Gal}(F_2/F_1)| = |F_2 : F_2^{\text{Gal}(F_2/F_1)}|$, by using hypothesis and the fixed field theorem.
Now, suppose Gal$(F_2/F_1)$ is a normal subgroup of Gal$(F_2/F_0)$,
and take any $t \in F_2^{\text{Gal}(F_2/F_1)}$.
All conjugations of $t$ can be written as $\sigma (t)$, where $\sigma \in $ Gal$(F_2/F_0)$, since $F_2/F_0$ is a Galois extension.
I want to say $\sigma (t) \in F_2^{\text{Gal}(F_2/F_1)}$(unchanged by Gal($F_2/F_1$)).
Since Gal$(F_2/F_1)$ is a normal subgroup,
$\forall \tau \in \text{Gal}(F_2/F_1), \sigma^{-1} \tau \sigma \in \text{Gal}(F_2/F_1)$, so that
$\exists \tau' \in \text{Gal}(F_2/F_1), $ s.t. $\sigma^{-1} \tau \sigma = \tau'$.
Therefore $\sigma^{-1} \tau \sigma (t)= \tau' (t) = t$, so that
$ \tau \sigma (t)= \sigma (t)$.
Hence, $\sigma (t) \in F_2^{\text{Gal}(F_2/F_1)}$, so that $F_2^{\text{Gal}(F_2/F_1)}/F_0$ is a Galois extension.
Then, noting that $F_2^{\text{Gal}(F_2/F_1)}=F_1$, $F_1/F_0$ is a Galois extension. $\square$
thank you for Lubin.