Q: Give an example of a topological space having exactly 3 coverings up to equivalence (including a covering by the space itself).
Proof: There is a theorem that says that given a topological manifold $M$ with a universal cover, the coverings are determined up to isomorphism by the conjugacy classes of subgroups of $\pi_{1}(M)$ (i.e. Topological Manifolds by Lee, Thm 12.18)
So let's take a group with exactly three subgroups up to conjugacy, so a group with 1 proper subgroup up to conjugay. A bit of thought leads to taking $\mathbb{Z}_{4}$ since this is Abelian so no need to worry about conjugacy and has three subgroups, $\mathbb{Z}_{4}, \{0\}, \langle 2\rangle$. Hence there are exactly three coverings for any topological space with $\pi_{1}(M)=\mathbb{Z}_{4}$.
Hence we construct such a space as follows. Take a representation: $\mathbb{Z}_{4}=\langle a : a^{4}=1 \rangle$. Form the $1$-skeleton of a cell complex by taking the circle, and attach a disk by wraping the boundary $4$ times around the circle.
So first question, does this look alright? Second question, is there any way to imagine this thing? More precisely, can I embed this in $\mathbb{R}^{3}$? My first instinct says no, but I'm not sure.