Let $F_0\subset F_1\subset\cdots\subset F_n$ be a series of finite extensions of fields. Suppose that $E$ is an intermediate field of $F_n/F_0$. Is there necessarily a series of finite extensions $F=E_0\subset E_1\subset\cdots\subset E_m=E$ such that $$ \max_{1\le i\le m}[E_i:E_{i-1}]\le\max_{1\le i\le n}[F_i:F_{i-1}]?\tag{*} $$ Here we do not necessarily have $m=n$.
My question comes from the definition of iterated quadratic extensions: $F_n/F_0$ is an iterated quadratic extension if there exists $F_0\subset F_1\subset\cdots\subset F_n$ such that $[F_i:F_{i-1}]=2$ for each $i$. I was wondering if a subextension of an iterated quadratic extension is again a such kind of extension, so I would like to pose the question about $(*)$. A first thought would be taking $m=n$ and $E_i = F_i\cap E$, which turns out not working, as shown by the example given by user297024 (which he/she has deleted in the comments here): $$ F_0=\mathbb{Q},\qquad F_1=\mathbb{Q}(\sqrt{3}),\qquad F_2=\mathbb{Q}(\sqrt[4]{3}),\qquad F_3=\mathbb{Q}(\sqrt[8]{3}),\\ F_4=\mathbb{Q}(\sqrt{2},\sqrt[8]{3}), \qquad F_5=\mathbb{Q}(\sqrt{2},\sqrt{-1},\sqrt[8]{3}); \qquad E=\mathbb{Q}(\zeta_8\sqrt[8]{3}). $$ We have $E\subset F_5$ and $E\cap F_4=\mathbb{Q}(\sqrt{3})$, so $[E\cap F_5:E\cap F_4]=[E:\mathbb{Q}(\sqrt{3})]=4$, but $[F_i:F_{i-1}]=2$ for $i=1,2,3,4,5$. So can anything be said if I just need another series $F=E_0\subset E_1\subset\cdots\subset E_m=E$? Thank you for any help.
No. Consider $F_0 \subseteq F_1 \subseteq F_2 \subseteq F_3$ where $\mathrm{Gal}(F_3/F_0) \cong A_4$ and the subfields $F_0, F_1, F_2, F_3$ correspond to the subgroups $A_4, \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}, 1$. We see that $\max [F_i: F_{i - 1}] = 3$. On the other hand, if you let $E$ be the subfield corresponding to a subgroup of index $4$ in $A_4$, then there is no intermediate subgroup between $A_4$ and a subgroup of index $4$ (since $A_4$ has no subgroups of index $2$). It follows that the only chain of subfields from $F$ to $E$ is given by $F = E_0 \subseteq E_1 = E$ where $[E_1: E_0] = 4$.
For an explicit example of such $F_3/F_0$, you may take $F_3 = k(x_1, x_2, x_3, x_4)$, $F_0 = k(e_1, e_2, e_3, e_4, \Delta)$, where the $e_i$'s are the elementary symmetric polynomials in the $x_i$'s and $\Delta = \prod_{i < j} (x_i-x_j)$. You can also find such field extensions with $F_0 = \mathbb{Q}$ if you want.