Give an example of a noncyclic Abelian group all of whose proper subgroups are cyclic.

The simplest possible example of this would be $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, as this is abelian and is the smallest group which is not cyclic. It is also known as the Klein four-group.

The first example that came to mind, probably because I've spent so much time with it lately, is $\mathbb{Z}(p^{\infty})$, which is of course isomorphic to the group of all $p^n$-th roots of unity, $n=0,1,2, \ldots$. What I've always liked about this group is that all proper subgroups are finite as well as cyclic, while the group itself is infinite and non-cyclic. Plainly, the other examples are far simpler. Let this be a lesson to the OP: learn enough mathematics and you may easily overlook simple examples.

More generally, any finitely generated noncyclic abelian group whose subgroups are cyclic has the form $\mathbb{Z}_p \times \mathbb{Z}_p$, where $p$ is prime.

Indeed, each finitely generated abelian group $G$ has the form $\mathbb{Z}_{n_1}\times ... \times \mathbb{Z}_{n_r} \times \mathbb{Z}^n$ with $n_1 \ | \ n_2 \ | \ ... \ | \ n_r$ and $n_1>1$.

Case 1: $n=0$. $G$ has to be noncyclic so $r\geq 2$. There is exists a prime $p$ dividing each $n_i$ and either $\mathbb{Z}_p \times \mathbb{Z}_p$ is a proper noncyclic subgroup of $G$ or $G= \mathbb{Z}_p \times \mathbb{Z}_p$.

Case 2: $n,r \neq 0$. $\mathbb{Z}_{n_1} \times ... \times \mathbb{Z}_{n_r} \times m \mathbb{Z}$ is a proper noncyclic subgroup of $G$.

Case 3: $r=0$ and $n \neq 0$. $G$ has to be noncyclic so $n\geq 2$. So $\mathbb{Z} \times m \mathbb{Z}$ is a proper noncyclic subgroup of $G$.