Theorem 1   A group $G$ is locally cyclic iff every subgroup $H\le G$ is locally cyclic.

Proof. Let $G$ be a locally cyclic group and $H$ a subgroup of $G$ Let $S$ be a finite subset of $H$ Then the group $\langle S\rangle$ generated by $S$ is a cyclic subgroup of $G$ by assumption. Since every element $a$ of $\langle S\rangle$ is a product of elements or inverses of elements of $S$ and $S$ is a subset of group $H$ $a\in H$ Hence $\langle S\rangle$ is a cyclic subgroup of $H$ so $H$ is locally cyclic.

Conversely, suppose for every subgroup of $G$ is locally cyclic. Let $H$ be a subgroup generated by a finite subset of $G$ Since $H$ is locally cyclic, and $H$ itself is finitely generated, $H$ is cyclic, and therefore $G$ is locally cyclic.