This was on my final last semester (to find such a cover), and I missed it. Here are my thoughts on it since then:
I know that the universal cover of $X = \mathbb{R}P^2\vee\mathbb{R}P^2$ is (loosely) a string of 2-spheres that goes off in both directions.
I also know that the fundamental group of $X$ is the group with presentation $<a,b|a^2=b^2=1>$. Any covering space of $X$ will have a fundamental group that is (isomorphically speaking) a subgroup of this group. Furthermore an irregular covering space will have fundamental group that is isomorphic to a non-normal subgroup of $\pi_1(X)$. $\mathbb{Z}_2$ is such a group.
Now I want to find a covering space of $X$ that has fundamental group $\mathbb{Z}_2$. This is where things get a little shaky for me. How do I find such a space. I know that the number of sheets will be infinite since the number of cosets of a finite subgroup in a infinite group is infinite.
So here's what I'm thinking. If I take my infinite string of 2-spheres and quotient one of them into an $\mathbb{R}P^2$ via the antipodal map (getting something like an $\mathbb{R}P^2$ with two strings of 2-spheres coming off a single point). I think this will do the trick. I know this is being a little sloppy (not explicitly giving the maps, but I hope it is obvious- If not I can clarify).
So I guess what I'm hoping is that someone can help me figure out if this is right or wrong, but even more what I'm hoping for is that someone can share some insight into this problem (or this sort of problem in general)- maybe provide a better way to look at this problem.
Sorry if this was long winded.
Thanks much :)
Edit: I believe solution is also wrong. For any neighborhood of a point that maps to the point that is the wedge of $\mathbb{R}P^2$ and 2 $S^2$'s will not be homeomorphic to any neighborhood of the point where the two projective planes are joined. I believe Dylan has a correct solution, but I am still hoping for some insight.