Following the notation of Mas-Collel, Whinston, and Green, consider a family of budget sets $\mathcal{B}=\{\{x,y\},\{x,y,z\}\}$. To make the example concrete, let's let
- $x$ be a book
- $y$ be a left shoe
- $z$ be a right shoe
For motivation, let's suppose the individual prefers going for a walk over reading a book. However, to go for a walk, he needs two shoes. If he can only have one shoe, then he would rather choose the book instead.
Thus when faced with Budget set $\{x,y\}$, the consumer chooses the the book: $C(\{x,y\})=\{x\}$. However when faced with all three alternatives, he chooses the shoes: $C(\{x,y,z\})=\{y,z\}$.
As we constructed it, this choice rule $C$ violates the Weak Axiom of Revealed Preference:
Suppose there exists some budget set $B_1 \in \mathcal{B}$ with $x,y \in B_1$ and $x \in C(B_1)$. Then for any $B_2 \in \mathcal{B}$ with $x,y \in B_2$ and $y\in C(B_2)$, we must also have $x \in C(B_2)$.
In our case, since $C(\{x,y\})=x$, then WARP would require $x \in C(\{x,y,z\})$ since $y \in C(\{x,y,z\})$. However, it seems we were able to motivate a sensible reason why $C(\{x,y,z\})=\{y,z\}$.
Side note: I have left prices of the goods out of this story. (Hopefully this is acceptable; to be fair, MWG's chapter on choice rules doesn't mention prices either. (I assume we should imagine the price of these goods are all 0, or all the same price -- would that be the right way to look at it?)
This example brings me to the question posed in the title: Can we have sensible choice behavior that violates WARP?