3-set-lemma in (naive) set theory By accident I came across the following curious, very general statement:

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 $i \in \{1,2,3\}$.

A simple example shows that there cannot be a "2-set-lemma" : let $X = \{0,1,2\}$ and let $f$ be the fixpoint-free bijection with $0\mapsto 1\mapsto 2\mapsto 0$. 
But I can't prove (or refute) the statement highlighted above. Can anyone give me a hint?
I would also be glad for a reference, where did this statement appear, does it have a proper name?
 A: Smells like Zorn's lemma.
Consider the set $\mathcal T$ of all tuples $(X_1,X_2,X_3)$ where the $X_i$ are


*

*pairwise disjoint subsets of $X$ and

*$f(X_i)\cap X_i=\emptyset$ and

*$f(X_1\cup X_2\cup X_3)\subseteq X_1\cup X_2\cup X_3$.


For example, $(\emptyset,\emptyset,\emptyset)\in\mathcal T$.
We say $(X_1,X_2,X_3)\le(Y_1,Y_2,Y_3)$ if $X_i\subseteq Y_i$ for $i=1,2,3$.
Given a chain $(X_1^{j},X_2^{j},X_3^{j})$ in $\mathcal T$ for some index set $J$, let $Z_i=\bigcup_{j\in J} X_i^j$. Then $(Z_1,Z_2,Z_3)\in\mathcal T$ because any conflict with the defining conditions would already occur at some $(X_1^{j},X_2^{j},X_3^{j})$. Hence we can apply Zorn's lemma and find a maximal triple $(X_1,X_2,X_3)$ in $\mathcal T$. 
Suppose $a\in X\setminus(X_1\cup X_2\cup X_3)$. Then $a\notin f(X_1\cup X_2\cup X_3)$. Let $a_0=a$ and recursively $a_{n+1}=f(a_n)$. 
Assume first $a_n\notin X_1\cup X_2\cup X_3$ for all $n$.
If the sequence $a_n$ is injective, add the $a_n$ suitably to the $X_i$, e.g., 
$$\begin{align}Y_1&=X_1\cup\{\,a_n: n\text{ odd}\,\},\\Y_2&=X_2\cup\{\,a_n: n\text{ even}\,\},\\Y_3&=X_3\end{align}$$ then $(Y_1,Y_2,Y_3)>(X_1,X_2,X_3)$, contradicting maximality.
So we may assume the sequence is not injective, hence it is eventually periodic with some period length $p>1$. If $p$ is even, the construction above still works.
If $p$ is odd, then $p\ge 3$ and we let
$$\begin{align}
Y_1&=X_1\cup \{\,a_n: p\mid n\,\},\\Y_2&=X_2\cup \{\,a_n: n\bmod p\in\{1,3,5,7,\ldots\}\,\},\\
Y_3&=X_3\cup \{\,a_n: n\bmod p\in\{2,4,6,8,\ldots\}\,\}.\\
\end{align}$$
Again, $(Y_1,Y_2,Y_3)>(X_1,X_2,X_3)$. 
Remains the case that $a_n\in X_1\cup X_2\cup X_3$ for some $n$, and hence for all $n\ge $ some minimal $N$, where clearly $N>0$. 
By symmetry we may assume wlog that $a_N\in X_3$. 
This time 
$$\begin{align}
Y_1&=X_1\cup \{\,a_n: n<N, n\text{ odd}\,\},\\
Y_2&=X_2\cup \{\,a_n: n<N, n\text{ even}\,\},\\
Y_3&=X_3\\
\end{align}$$
works.
