My approach: If $\mathbb{H}$, the upper half plane in $\mathbb{C}$ were the universal cover of $\Sigma$, then $\pi_1{\Sigma} \subset PSL(2,R)$, is a discrete subgroup which is not possible, as $\mathbb{Z} \bigoplus \mathbb{Z}$ cannot be contained in PSL(2,R) as a discrete subgroup. But I don't know how to prove the later?
Any help is appreciated! Thanks
You would have two matrices $A$, $B\in\text{SL}_2(\Bbb R)$ of infinite order which compute up to sign. That means their squares commute, so we may assume they commute. Then one can show that $B$ must be a linear combination of $A$ and $I$. The condition that $B$ has determinant $1$ will mean that $A$ and $B$ will lie in a one-parameter subgroup, so discreteness will finish off the job.