There are two questions here, an explicit one, and another (more vague) one that motivates it:

I am pretty certain the following should have a negative answer, but at the moment I'm not seeing how to argue about this and cannot locate an appropriate reference.

In set theory without choice, suppose $X$ is an infinite set such that for every positive integer $n$, we can split $X$ into $n$ (disjoint) infinite sets. Does it follow that $X$ can be split into infinitely many infinite sets? What would be a reasonably weak additional assumption to ensure the conclusion.

("Reasonably weak" would ideally be something that by itself does not suffice to give us that $X$ admits such a splitting, but I am flexible.)

This was motivated by a question at Math.SE, namely whether an infinite set can be partitioned into infinitely many infinite sets. This is of course trivial with choice. In fact, all we need to split $X$ is that it can be mapped surjectively onto ${\mathbb N}$.

However, without choice there may be counterexamples: A set $X$ is *amorphous* iff any subset of $X$ is either finite or else its complement in $X$ is finite. It is consistent that there are infinite amorphous sets. If $X$ is infinite and a finite union of amorphous sets, then $X$ is a counterexample. The question is a baby step towards trying to understand the nature of other counterexamples.

Note that any counterexample must be an infinite Dedekind finite (iDf) set $X$. One can show that for any iDf $X$, ${\mathcal P}^2(X)$ is Dedekind infinite. For any $Y$, if ${\mathcal P}(Y)$ is Dedekind infinite, then $Y$ can be mapped onto $\omega$ (this is a result of Kuratowski, it appears in pages 94, 95 of Alfred Tarski, "Sur les ensembles finis", Fundamenta Mathematicae 6 (1924), 45–95). As mentioned above, our counterexample $X$ cannot be mapped onto $\omega$, so ${\mathcal P}(X)$ must also be an iDf set.

The second, more vague, question asks what additional conditions should a counterexample satisfy.

1more comment