Let $L/K$ be an algebraic field extension. It is clear that $L$ can be written as a union of a directed family of finite subextensions, namely
$$
L = \bigcup_{x \in L} K(x).
$$
When is it possible to choose a totally ordered family of finite subfields $(L_i)_{i\in I}$, $L_j \subseteq L_k$ iff $j \leq k$, satisfying
$$
L = \bigcup_{i\in I} L_i
$$
and when can $I$ be chosen as a countable set, i.e. $I=\mathbb N$?
For $K=\mathbb Q$ and its algebraic closure $L= \overline{\mathbb{Q}}$ it should work because $\mathbb Q $ is countable. We can choose a bijective function $f:\mathbb N \rightarrow \overline{\mathbb Q}$ to define $x_n = f(n)$ and $L_n= K(x_1, \ldots , x_n)$.
Is there a general criterion when $K$ is a local field? For example an arithmetically profinite extension $L/K$ admits the choice of a countable family of subfields $(L_n)_{n\in \mathbb N}$.