Does every infinite field contain a countably infinite subfield?

Yes. If $S$ is a countably infinite subset of $K$ then the subfield generated by $S$ is countable.

Since the question got six upvotes before any answers, maybe this is not obvious. Details:

Let $S_0=S$. Let $S_{n+1}$ consist of the elements of $S_n$ together with all the $x+y$, $x-y$, $xy$ and $x^{-1}$ for $x,y\in S_n$. Then $\bigcup_{n=0}^\infty S_n$ is a countable subfield.

(Yes, the same argument shows that $K$ has a subfield of any infinite cardinality less than or equal to the cardinality of $K$.)

As Ulrich points out, any field generated by a countable set of elements of $K$ will be countable, but this doesn't guarantee a proper subfield, so let me pose a different question: does every infinite field properly contain an infinite subfield? (I'll even throw out the word countable.) Over course $\Bbb Q$ is a trivial counterexample in characteristic $0$, so let's consider characteristic $p$.

The answer to that question is no.

We know the finite extensions of $\Bbb F_q$ are $\Bbb F_{q^n}$ for naturals $n$, which are unique for their sizes, with inclusions $\Bbb F_{q^d}\subseteq\Bbb F_{q^n}$ if and only if $d\mid n$ (i.e. $d$ is a divisor of $n$). But can we characterize all algebraic extensions of $\Bbb F_q$? Yes: to do so we need supernatural numbers for exponents.

A supernatural number is a formal infinite product $n=\prod_p p^e$ of prime powers, with no restrictions on the exponents, and in fact $e=\infty$ is an allowed exponent. The notion of divisibility for supernatural numbers should be obvious.

Define $\Bbb F_{q^n}$ for supernatural $n$ to be $\bigcup_{d\mid n}\Bbb F_{q^d}$ within $\overline{\Bbb F_{q}}$, where $d$ ranges over all finite, natural number divisors of $n$. Given any $K\subseteq\overline{\Bbb F_q}$, we can associate to it the supernatural number $n$ which is the LCM of all finite $d$ for which $\Bbb F_{q^d}\subseteq K$, or equivalently $n=\prod p^{e(p)}$ where $e(p)$ is maximal subject to $\Bbb F_{q^{p^e}}\subseteq K$ (or $e(p)=\infty$ if that holds true for all $e$).

Exercise: Prove $K=\Bbb F_{q^n}$ where $n$ is the supernatural number associated to $K$.

Indeed, the lattice of supernatural numbers ordered by divisibility is isomorphic to the lattice of algebraic extensions of $\Bbb F_q$ ordered by inclusion. The isomorphism is $n\mapsto \Bbb F_{q^n}$.

In particular, this means for any prime $\ell$, the infinite field $\Bbb F_{q^{\ell^\infty}}$ does not properly contain any infinite subfields. Contrast with $\overline{\Bbb F_q}$, which properly contains infinite subfields $\Bbb F_{p^n}$ precisely when $n$ is an infinite supernatural which properly divides $\prod p^\infty$. (In particular it properly contains the aforementioned infinite subfields $\Bbb F_{q^{\ell^\infty}}$s!)

There is also some Galois theory that can be discussed here. The fundamental theorem of Galois theory in this context says the Galois extensions of $\Bbb F_q$ are lattice anti-isomorphic to the lattice of closed subgroups of the absolute Galois group ${\rm Gal}(\overline{\Bbb F_q}/\Bbb F_q)\cong\widehat{\Bbb Z}\cong\prod_p\Bbb Z_p$ considered with the profinite topology (note $\Bbb Z_p$ is the additive group of $p$-adic integers, not to be confused with $\Bbb Z/p\Bbb Z$). The closed subgroups, enumerated by supernaturals $n=\prod_p p^{e}$, are of the form $\prod_p \Bbb Z_p/(p^e)$, where $\Bbb Z_p/(p^e)\cong\Bbb Z/p^e\Bbb Z$ with $0\le e<\infty$ and $\Bbb Z_p/(p^e):=\Bbb Z_p$ when $e=\infty$ (which makes sense because $p^e\to0$ in $\Bbb Z_p$'s topology as $e\to\infty$).

Suppose $K$ has characteristic $p$.

If $K$ has a transcendental element $\alpha$, then $\mathbf{F}_p(\alpha) \subseteq K$ is countably infinite, being isomorphic to the rational function field $\mathbf{F}_p(x)$.

Otherwise, $K \subseteq \overline{\mathbf{F}}_p$ which is countable, so you can just take $K$ itself to be the countably infinite subfield.