Find a non-abelian group whose proper subsets are all abelian.
Just a remark, $\displaystyle S_3$ is definitely a good choice, but why? It's a first-week-of-group-theory matter that every group of order at most five is abelian, thus if you can find a non-abelian group whose proper subgroups all have size at most five, then you're golden. But, clearly any group of order six has the property that every proper subgroup has order at most five, and so in particular, any non-abelian group of order six will have the property you seek. Ta-da, $\displaystyle S_3$!