All of the field extensions involved are assumed to be finite.
This lemma is from Isaacs' Algebra, serving as a key step toward the establishment of Galois' criterion of solvability. The proof of the original lemma is as shown in the pictures.
The original lemma: Suppose $F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$, where for $1 \leq i \leq r$, each of the extensions $F_{i-1} \subset F_i$ is Galois with an abelian Galois group. Suppose $F \subseteq E \subseteq L$ and $E$ is Galois over $F$. Then $\operatorname{Gal}(E/F)$ is solvable.
It seems to me that the lemma holds still if one replace all 'Galois' in the original version by 'normal'. Then it becomes like this:
The modified lemma: Suppose $F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$, where for $1 \leq i \leq r$, each of the extensions $F_{i-1} \subset F_i$ is normal with an abelian Galois group. Suppose $F \subseteq E \subseteq L$ and $E$ is normal over $F$. Then $\operatorname{Gal}(E/F)$ is solvable.
But I am not sure if the modified lemma am right. I also write down a proof for the version lemma assuming normality. If it will not take a lot of time, please verify the following proof.
Proof: We work by induction on $r$. Let $E_1=\langle F_1, E\rangle$. The fact that $E_1$ is normal over $F$ follows from normality of $F \subseteq E$ and $F \subseteq F_1$. Thus, $F_1 \subseteq E_1$ is also normal. We have a tower of length $r-1$ of normal extensions from $F_1$ to $L$, and by the inductive hypothesis one concludes $\operatorname{Gal}(E_1/F_1)$ is solvable.
Consider the tower $F \subset F_1 \subset E_1$. Since $F \subset F_1$ and $F \subset E_1$ are normal, $\operatorname{Gal}(E_1/F_1) \lhd \operatorname{Gal}(E_1/F)$, and $\operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/F_1) \cong \operatorname{Gal}(F_1/F)$. Given both $\operatorname{Gal}(E_1/F_1)$ and $\operatorname{Gal}(F_1/F)$ are solvable, one knows $\operatorname{Gal}(E_1/F)$ is solvable.
Next consider the tower $F \subset E \subset E_1$. Again we have two normal extensions $F \subset E$ and $F \subset E_1$, which follows $\operatorname{Gal}(E_1/E) \lhd \operatorname{Gal}(E_1/F)$, and $\operatorname{Gal}(E/F) \cong \operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/E)$. One deduce $\operatorname{Gal}(E/F)$ is solvable as a quotient of a solvable group.