Does ZF prove that there exists a set S such that S is not in the closure of {{s} : s in S} under at-most-countable unions?



Can iterating countable unions give every set? (ZF)
Moti Gitik proved (assuming large cardinals) that all uncountable alephs can have cofinality ω. [*All uncountable cardinals can be singular*, Israel J. Math. 35 (1980), 61–88] I believe this may be the model you're looking for, but I don't know what happens to non-wellorderable sets in that model. According to the abstract copied below, this model is very close to what you have in mind.



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> Assuming the consistency of the existence of arbitrarily large strongly compact cardinals, we prove the consistency with ZF of the statement that every infinite set is a countable union of sets of smaller cardinality. Some other statements related to this one are investigated too.
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Every wellorderable set $S$ in this model belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions. The proof is by induction on the cardinality of the infinite set $S$. As in the abstract, we may write $S = \bigcup\_{i<\omega} S\_i$ where $|S\_i| < |S|$ for each $i$. By the induction hypothesis, each $S\_i$ belongs to the closure of $\{\{s\}:s \in S\}$ under iterated countable unions, and thus $S$ belongs to this closure too.




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Actually, every set $S$ in this model belongs to the closure of $\{\{s\} : s \in S\}$ under iterated countable unions; this is essentially what Gitik's Theorem 6.3 says.


