If $F$ is a countable field, then proving that $F$ has algebraic closure is quite simple: there can be at most countable number of monic irreducible polynomials over $F$, let they form the set $\mathcal{L}$. For each polynomial in $\mathcal{L}$, we get an algebraic extension. Writing $\mathcal{L}$ as countable ascending union of finite subsets, one can obtain algebraic closure of $F$.
Of course the argument will not work if $\mathcal{L}$ is uncountable, some variation in argument is required. This already has been appeared in some other questions on mathstack.
But this raised one question to me:
Question: Does there exists a field $F$ with the following three properties:
(1) $F$ is uncountable
(2) $F$ is not algebraically closed.
(3) The number of monic irreducible polynomials of degree $>1$ is countable.
Note that $\mathbb{R}$ satisfies (1) and (2) but not (3).