Comb space has no simply connected cover

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.

I believe that I found an answer to my problem. The earlier approach of playing with open covers didn't pan out because I couldn't extend local injectivity of $p$ on a bunch of open sets to injectivity on their union.

But I found the following theorem, which solves the problem quite neatly. I found it in [1], along with a sketch proof. (But I'll prove it fully):

Theorem: Let $E$ be any space and $F$ be a metric space. Let $K\subset E$ be compact, $f:E\to F$ be continuous, locally injective near every point $p\in K$, and (globally) injective on $K$. Then $f$ is also injective on some open neighbourhood $N$ of $K$.

Proof: Let $g:E\times E\to \mathbb{R}$ be the function defined by $g(x,x')=d(f(x),f(x'))$. Around each point of the form $(k,k)\in K\times K$, there is an open product neighbourhood $V_k \times V_k$ such that $(V_k\times V_k)\cap g^{-1}(0) \subset \{ (x,x)\in E\times E \}$ (by local injectivity). Let $V=\cup_k \left(V_k\times V_k\right)$. Then the compact set $K\times K$ is contained in the open set $g^{-1}(0)^c \cup V$. By the generalized tube lemma, there are open sets $W,W'$ of $E$ so that $K\times K\subset W\times W'\subset g^{-1}(0)^c \cup V$. Then let $N=W\cap W'$. $\square$

Applying this theorem to $p:\tilde{X} \to X$, with $K$ being a lift of the left edge of $X$, gives an open neighbourhood of that lift which maps homeomorphically to an open neighbourhood of the left edge of $X$. Then we use the bad local nature of $X$ to find a loop in $X$ which lifts to a loop in $\tilde{X}$ which is not contractible. (Compactness is used again to find an open set which contains the entirety of a vertical line)

[1] Wagner,M., "Existence of schlicht integral manifolds for ordinary differential equations", Archiv der Mathematik, Volume 61, Number 6, 529-542 (1993)