Let a choice function be defined as a function $C:2^X \setminus \{\emptyset\} \rightarrow 2^X$ such that $C(A) \subseteq A$ for all $A \subseteq X$. Here, $2^X$ denotes the power set of $X$. We say a choice function is path independent if for all $A, B \in 2^X \setminus \{\emptyset\}$, $C(A \cup B) = C(C(A) \cup C(B))$. We say a choice function is rationalizable if there exists a binary relation $\succeq$ on $X$ that is complete and transitive such that $C(A) = C_{\succeq}(A)$ for all $A \subseteq X$ where $C_{\succeq}(A) = \{x \in A : x \succeq y \text{ for all } y \in A\}$. A choice function is nonempty if $C(A) \neq \emptyset$ for all nonempty $A \subseteq X$.
Construct a nonempty choice rule that is path independent but not rationalizable.
Let $X = \{x,y,z\}$. For all two-element subsets $A$ of $X$, let $c(A) = A$. Let $c(X) = \{x,y\}$. Path independence follows by construction, failure of rationality follows from the existence of a 2-cycle: $x \succ_c z$ $\&$ $z \succsim_c x$.