Universal Cover of $\mathbb{R}P^{2}$ minus a point I've already calculated that the fundamental group of $\mathbb{R}P^{2}$ minus a point as $\mathbb{Z}$ since we can think of real projected space as an oriented unit square, and puncturing it we can show it is homotopy equivalent to the boundary which is homotopy equivalent to the unit circle.

I am having some trouble finding the universal cover of punctured real projective space, however. I'm not sure how to construct one. I know that punctured real projective space is homeomorphic to the mobius band, but I'm not sure if that helps me here. Thank you for any help.

Letting $M$ denote the Mobius band, there is a universal covering map $\mathbb{R}^2 \to M$ with infinite cyclic deck transformation group $\langle r \rangle$ generated by the "glide reflection"
$$r : (x,y) \to (x+1,-y)
$$
As with any universal covering map, $M$ is the orbit space of this action. One can verify this by noticing that the vertical strip $[0,1] \times \mathbb{R}$ is a fundamental domain for $r$, its left side $0 \times \mathbb{R}$ being glued to its right side $1 \times \mathbb{R}$ by identifying $(0,y)$ to $(1,-y)$. This is exactly the gluing pattern for constructing the Mobius band that we all learned on our mother's knee.
