I have been trying to find a counterexample that all groups (finite and infinite) have an abelian subgroup (besides the trivial subgroup).
Any ideas?
Every non-abelian group has a non-trivial abelian subgroup: Let $G$ be a nonabelian group and $x\in G$, $x$ not the identity. Then $\langle x\rangle$ is an abelian subgroup of $G$.
EDIT: In case you are curious, there are nonabelian groups such that the only abelian subgroups are the cyclic ones generated by one element. For example, $S_3$ is a nonabelian group such that only the cyclic subgroups are abelian.
It's impossible, as the subgroup generated by one element is cyclic and a cyclic group is abelian.