The number $3$ plays an interesting role in the following statement:

$\newcommand{\S}{\sf(S_3)}\S$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$.

There are easy examples showing that one cannot get by using $2$ subsets only. Statement $\S$ can be proved using the axiom of choice. (Although the full force of (AC) is not needed as the strictly weaker Boolean Prime Ideal Theorem (BPI) is sufficient, but it appears that the $\S$ is not a theorem of $\ZF$.)

The motivation of this question is to find a weakening of $\S$ that is a theorem of $\newcommand{\ZF}{{\sf (ZF)}}\ZF$.

Question. Is the following statement a theorem of $\ZF$?

If $f:\mathbb{N}\to\mathbb{N}$ is fixed-point free, then there is a finite set ${\frak S}\subseteq {\cal P}(\mathbb{N})$ with $\bigcup{\cal S} = \mathbb{N}$ and $S\in {\cal S}$ with $S\cap f(S) = \emptyset$.