Consider the following
Theorem: Any algebraic field extension $K|F$ of infinite degree contains finite subextensions of arbitrarily high degree.
Proof: We'll prove that, for any n, there's a finite subextension $F_n|F$ of degree $\geq 2^n$, by induction on n.
The case $n=1$ is obvious. Suppose, by inductive hypothesis, that we have a finite subextension $F_{n-1}|F$ of degree $\geq 2^{n-1}$. Since $K|F$ is an infinite extension, there's an element $x\in K\setminus F_{n-1}$, which is algebraic over $F$; then $F_{n-1}(x)|F$ is a finite extension of degree $\geq 2^n$, as we wanted.
Does this rely on countable choice? Note that for each fixed $n$, we have only made a finite number of choices to get the desired subextension.
No, this is really just induction with finitely many choices.
The axioms of choice would come in if you would have wanted to say there is a countably infinite set of algebraically independent elements. With this proof, you would invoke Dependent Choice, which is stronger than countable choice. But I reckon the proof can be slightly modified to use just countable choice instead.