Without the Axiom of Choice, does every infinite field contain a countably infinite subfield? Earlier today I asked whether every infinite field contains a countably infinite subfield. That question quickly received several positive answers, but the question of whether those answers use the Axiom of Choice has spawned off an interesting discussion in its own right. Thus I'd like to pose that question separately:

Without using the Axiom of Choice, can it be shown that every infinite field contains a countably infinite subfield?

Note that the corresponding question for sets has a negative answer, but since fields have so much more structure than sets, it is not a priori unreasonable to me that there might exist an AC-less resolution for fields.

EDIT: For the purposes of this question, an infinite set is one that is not bijective with $\{1, \dots, n\}$ for any $n \in \Bbb N$. Thanks to @MartinSleziak for bringing this issue up in chat.

Impossible, by the result from Hickman (1978). He proves that the existence of a Dedekind-finite infinite field (that is, an infinite field does not have any countable subset) is consistent with ZF. More precisely, he proves the following theorem:

Theorem. For a prime $p$ and a sequence of natural numbers $m_0\mid m_1\mid\cdots$, it is consistent with ZF that there is an Dedekind-finite infinite field, such that the field is a union of a sequence of increasing finite fields $(F_n)_{n<\omega}$ with $|F_k| = p^{m_k}$.

(His original paper uses the terminology "medial" to denote a Dedekind-finite infinite set.) 
Furthermore, he proves that every Dedekind-finite infinite rings with only a finite number of zero-divisors is a field. Thus every Dedekind-finite infinite ring is a field or has infinite number of zero-divisors.
You can check the details in his paper:

Hickman, J. L. "Dedekind-finite fields." Bulletin of the Australian Mathematical Society 19.01 (1978): 117-124.

