 Solution

Not a topology. For every integer $k\ge 2$, the set $U_k=\mathbb{Z}\setminus k\mathbb{Z}$ is periodic with period $k$, so belongs to the family. Their union is $\mathbb{Z}\setminus \{0\}$: for a nonzero $n$, choose $k>|n|$, which cannot divide $n$. But this union has no positive integer period. If it had period $k$, membership of $-k$ would force membership of zero. Again arbitrary unions fail the topology axiom.