I'm having trouble solving the following exercise in Hatcher's Algebraic Topology(1.3 #5):
"Let $X$ be the subspace of $R^2$ consisting of the four sides of the square $[0,1] \times [0,1]$ together with the segments of the vertical lines $x=\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ inside the square. Show that for every covering space $\tilde{X}\to X$ there is some neighbourhood of the left edge of $X$ that lifts homeomorphically to $\tilde{X}$. Deduce that $X$ has no simply-connected cover."
I tried piecing together open sets in $\tilde{X}$ which were homeomorphic to open square shaped $\epsilon$-neighbourhoods of points in the left edge of $X$, but I couldn't glue them together coherently.
My second idea was to use the path lifting property to lift the left edge, but I couldn't extend it to a lift of an open neighbourhood of the left edge.
Edit: Some more detail on my first attempt:
Suppose I have two open squares $U_1$ and $U_2$ in $X$ (containing the left edge) with nonempty intersection, and that have preimages $\coprod_i U_{1,i}$ and $\coprod_j U_{2,j}$ with each piece of that mapping homeomorphically onto $U_1$ and $U_2$ respectively.
Suppose I've chosen a specific $U_{1,i}$, and I want to choose a specific $U_{2,j}$ so that their union maps homeomorphically to $U_1$ union $U_2$. Such a choice can be specified by a point in the chosen $U_{1,i}$ that maps into $U_2$, but for each such point, there could be a different choice of $j$.
My intution tells me that to make this choice, one needs to appeal to the bad local nature of the left edge of $X$, (so that you're certain to get some of the vertical lines inside of the desired open set) but I'm not sure how to proceed with that.