Here it is (deleted now):
Let's define $V=\{u_1^{k_1}u_2^{k_2}...u_n^{k_n}:n \in \mathbb N,k_i\in \mathbb Z, u_i\in U\},u_i$'s need not be distinct.
**Showing that $V$ is the 'smallest' subgroup that contains $U$:**
By choosing $n=1$, it follows that $V$ contains $U$.
$V$ is clearly closed under group operation.
For any $u\in U,$ choosing $n=1$ and $k_1=-1$, it follows that $u^{-1}\in V$. So $V$ is closed under inverses.
$V$ contains identity of the group. It follows that $V$ is a subgroup.