Is a weak version of the three sets Lemma provable in ZF?

Question

The Three Sets Lemma is the following Lemma:

Lemma: Let $f(x)$ be a function from $X$ to $X$ where $f(x)$ has no fixed points. Then there exists a partition of $X$ into three disjoint sets $X_1$, $X_2$, $X_3$ such that for any $i$, with $1 \leq i \leq 3$, we have $X_i \cap f(X_i)= \emptyset$.

The Three Sets Lemma is a theorem of ZFC. See e.g. Hagen von Eitzen's answer to 3-set-lemma in (naive) set theory. However, the Three Sets Lemma is not a theorem of ZF, although it is, much weaker than choice, and is in fact even weaker than the Boolean Prime Ideal theorem. See this other Mathoverflow question and the answer by godelian.

Let $S(k)$ be the following statement: "Let $f(x)$ be a function from $X$ to $X$ where $f(x)$ has no fixed points. Then there exists a partition of $X$ into $k$ disjoint sets $X_1, X_2, \dotsc X_k$ such that for any $i$, with $1 \leq i \leq k$, we have $X_i \cap f(X_i)= \emptyset$.

Note that $S(3)$ is then the Three Sets Lemma (and $S(1)$ and $S(2)$ are both just false).

Question: Is there any $k$ such that $S(k)$ is a theorem of ZF?

I suspect that answer is no.

I also suspect that the even weaker statement where we swap the order of the quantifiers (so first choose an $f(x)$ and then are allowed to choose $k$) is not a theorem of ZF.

Note that there is this other weaker variant of the Three Sets Lemma which is a theorem of ZF, but it requires the assumption that the set $X$ is well-ordered.

Answer 1

The answer is indeed no. Consider a Russell sequence $A_i,i\in\mathbb N$, that is a collection of disjoint two-element sets such that no infinite subfamily of them has a choice function. It is consistent with ZF that such a sequenc exists. Let $X=\bigcup_iA_i$ and let $f:X\to X$ be a function swapping elements of each $A_i$. Clearly $f$ has no fixed points. But any subset $Y\subseteq A_i$ such that $Y\cap f(Y)=\varnothing$ is finite - it must contain at most one element from each $A_i$, and had it been infinite, we could use it to produce a choice function for infinitely many of them. In particular, there is no finite partition of $X$ into such sets.