If we have $F_n\geq\ldots\geq F_1$ fields such that $F_{i+1}$ is a separable splitting field over $F_i$ for all $i=1,\ldots,n-1$ is it true that $F_n$ is a splitting field over $F_1$? I think I can prove it for the finite case, but am unsure about the infinite case. For the finite case here's what I had. For simplicity we'll assume characteristic $0$, so everything is separable.
$\textbf{Proof:}$ If $n=1$ there is nothing to show. Suppose $n>1$. Let $$\sigma:F_{n-1}\to F_{n-1}$$ be an automorphism leaving $F_1$ fixed. Since $F_n$ is a splitting field over $F_{n-1}$ it follows that we can extend $\sigma$ in $[F_n:F_{n-1}]$ ways to an automorphisms of $F_n.$ If $F_{n-1}$ were a splitting field over $F_1,$ then there would be $[F_{n-1}:F_{n-1}]$ such $\sigma,$ hence there would be at least $$[F_{n}:F_{n-1}][F_{n-1}:F_1]=[F_n:F_{n-1}]$$ automorphisms of $F_n$ leaving $F_1$ fixed, and $F_n$ would be a splitting field.
If $F_{n-1}$ is not a splitting field over $F_{1}$, then applying the contrapositive of the above to $F_{n-1}$ it would follow that $F_{n-2}$ were not a splitting field over $F_1,$ so $n-2>1.$ By infinite descent $F_{n-1}$ is a splitting field over $F_1,$ and so $F_n$ is a splitting field over $F_1.\blacksquare$
The above argument hinges on the fact that $F_n$ has $[F_{n}:F_1]<\infty$ isomorphisms into a subfield of $\overline{F_n}$ leaving $F_1$ fixed, hence since we've shown that many automorphisms every isomorphisms of that type is an automorphism. This argument cannot apply to the infinite case, for even if we have $\infty$ automorphisms we cannot apply this counting argument. Is there a more general way to make it work for the infinite case?