TITLE: Uncountable subset with uncountable complement, without the Axiom of Choice
QUESTION [13 upvotes]: Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is countable, then $\mathcal{A}(X)$ coincides with the power set $\mathcal{P}(X)$ of $X$. Now suppose that $X$ is uncountable. Assuming the axiom of choice, we can conclude that $\mathcal{A}(X) \ne \mathcal{P}(X)$, since $|X| = |X| + |X|$. So the question is:

Can we prove in ZF that $\mathcal{A}(X) \ne \mathcal{P}(X)$ for every uncountable set $X$?

I'm assuming that a set $X$ is uncountable if there is no injective function $f : X \rightarrow \mathbb{N}$.

REPLY [10 votes]: I should add on Andres' answer with a minor variation:
It is consistent with ZF without the axiom of choice that there is a set $X$ such that every subset of $X$ is countable or co-countable. These sets are called $\aleph_1$-amorphous (and sometimes quasi-minimal). They make peculiar counterexamples to some propositions.
If $X$ is $\aleph_1$-amorphous then $\mathcal P(X)=\mathcal A(X)$. Note that this is the "maximal" counterexample that we can produce since anything larger would have a subset which is not countable nor co-countable.
One remark is that amorphous sets are vacuously $\aleph_1$-amorphous, since every subset is finite (ergo countable) or co-finite (ergo co-countable). However it is possible to have an $\aleph_1$-amorphous set and still retain the Principle of Dependent Choice, a choice principle which amongst other things imply that every infinite set has a countable subset. In such model the $\aleph_1$-amorphous set is "non-degenerate", namely it has a countably infinite subset.