 Solution

Since $U_m\cap U_n=U_{\gcd (m,n)}$ when the greatest common divisor is at least $2$, and is empty otherwise, finite intersections of the $U_n$ are again among the allowed unions. Arbitrary unions are built into the definition; the empty union is empty, and $X={\bigcup }_{n\in X}{U}_n$. Thus these sets form a topology.

The space is not Hausdorff: every neighborhood of $4$ contains its divisor $2$, and every neighborhood of $2$ contains $2$, so $2$ and $4$ cannot have disjoint neighborhoods.