= Solution
The assertion needs \b[$E$ connected]. Without that assumption, the disjoint union of two tori mapping by the identity on each component is a covering of the <torus> and is none of the three listed connected spaces. The correct general statement classifies each connected component; here is the intended connected proof.
The <universal cover> is $\mathbb R^2\to\mathbb R^2/\mathbb Z^2=T$, and $\pi_1(T)=\mathbb Z^2$. The classification of connected coverings of a connected, locally path-connected, semilocally <simply connected> space identifies $E$ with $\mathbb R^2/H$ for a <subgroup> $H\le\mathbb Z^2$, acting by translations. A <subgroup> of $\mathbb Z^2$ is free abelian of rank zero, one or two. For rank zero the quotient is the plane. For rank one, a linear coordinate change sends its generator to a translation in one coordinate, yielding $S^1\times\mathbb R$. For rank two, a linear coordinate change sends a lattice basis to the standard basis, yielding a <torus>. Thus \b[a connected covering is a plane, cylinder or <torus>]. For arbitrary $E$, its covering components are disjoint unions of these types, rather than necessarily one such space.
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