 Solution

The square gives one vertex, edges $a,b$, and one 2-cell attached by $ab{a}^{-1}b$, hence ${\pi }_1(K)=⟨a,b\mid ab{a}^{-1}=b^{-1}⟩$. Based connected covers correspond contravariantly to subgroups of ${\pi }_1$. The subgroup $⟨a^2,b⟩\simeq {\mathbb{Z}}^2$ gives the two-sheeted torus cover; $⟨b⟩\simeq \mathbb{Z}$ gives an infinite cylindrical cover. If the total space is compact, fibres over a point are compact discrete and hence finite, so the subgroup has finite index. Conversely a finite-index cover has finitely many compact lifted cells and is compact.