Space without universal cover (Solution Exercise A. Hatcher 1.3.5) 
Problem:
Let $X$ be the subspace of $R^2$ consisting of the four sides of the square $[0, 1]×[0, 1]$
together with the segments of the vertical lines x = 1/2, 1/3, 1/4, ··· inside the square.
Show that for every covering space $p:\tilde{X}\to X$ there is some neighborhood of the left
edge of $X$ that lifts homeomorphically to $\tilde{X}$ . Deduce that $X$ has no simply-connected
covering space.

Proved until now:

*

*$p$ is suryective as $\tilde{X}$ is simply conected it implies conected an it imlies that $|p^{-1}(x)|$ is constant for all $x \in X$ so if a point wouldn't have any preimage then no point would have.


*$X$ is path conected then it is conected.

The title is because it is a space tha is semi-locally simply conected but has not univesal cover, so it make a interesting example. I have read the ideas in other questions in here and i dont get the details, for example in here Solution, it takes a evenly cover of a set (the left edge) that could not be a distingued neighborhood or at least i dont see why it is.
 A: Funnily enough, I also worked on this problem from Hatcher quite recently, and I also found it difficult to follow the arguments online - particularly in places where various lifts defined on open sets are claimed to join correctly. I'll share my approach, which looks different to the other solutions online in that I start off by arranging my open cover of $X$ to look nice and simple. These manipulations are probably overkill, but for me personally, I find that they make it easier for me to reason about the way that my lifts join together.

Since $p: \tilde X \to X$ is a covering, there exists a collection of open subsets $\{ U_\alpha \}$ covering $X$ such that for each $\alpha$, $p^{-1}(U_\alpha)$ is a disjoint union of open sets each mapped homeomorphically to $U_\alpha$ by $p$.
Each $U_\alpha$ is a union of sets of the form $R \cap X$ where each $R$ is an open rectangle in $\mathbb R^2$. (This is true by the definition of the subspace topology on $X$ induced from $\mathbb R^2$ and by definition of the product topology on $\mathbb R^2$.) So without loss of generality, we can simply assume that all of the $U_\alpha$ is of the form $R_\alpha \cap X$, where each $R_\alpha$ is an open rectangle.
Since the left hand edge of $X$ is compact, it is covered by finitely many such $U_\alpha$'s.
Reordering the $U_\alpha$'s (and possibly discarding some of them), we find that the left hand edge of $X$ is covered by overlapping open sets of the form
$$ U_1 = \left([0, \epsilon_1) \times [0, b_1) \right) \cap X$$
$$ U_2 = \left([0, \epsilon_2) \times (a_2, b_2) \right) \cap X$$
$$ U_3 = \left([0, \epsilon_3) \times (a_3, b_3) \right) \cap X$$
$$ \dots $$
$$ U_N = \left([0, \epsilon_N) \times (a_N, 1] \right) \cap X$$
where
$$ 0 < a_2 < b_1 < a_3 < b_ 2 < \dots < a_N < b_{N-1} < 1 $$
and where each $p^{-1}(U_n)$ is a union of disjoint open sets mapped homeomorphically onto $U_n$ by $p$.

Our next objective is to construct a lift of a neighbourhood of the left-hand edge of $X$, where the neighbourhood is of the form $\left([0, c) \times [0, 1] \right) \cap X$ for some $c > 0$. We'll do this in $N$ steps.
Step 1. Write $p^{-1}(U_1) = \coprod_i \tilde U_{1, i}$, where each $\tilde U_{1, i}$ is mapped homeomorphically to $U_1$ by $p$. Pick any $\tilde U_{1, i}$. We can construct a lift of $U_1$ to $\tilde X$ using the homeomorphism $(p |_{\tilde U_{1, i}})^{-1} : U_1 \to \tilde U_{1, i}$.
Thus we have constructed a map $f_1$ that lifts $([0, c_1) \times [0, b_1)) \cap X$ to homeomorphically into $\tilde X$, where $c_1 = \epsilon_1$.
Step 2. Write $p^{-1}(U_2) = \coprod_i \tilde U_{2, i}$, where each $\tilde U_{2, i}$ is mapped homeomorphically to $U_2$ by $p$. Pick any $t \in (a_2, b_1)$. One of these $\tilde U_{2, i}$'s contains the image of $(0, t)$ under $f_1$. This $\tilde U_{2, i}$ then also contains the image of $([0, c_2) \times \{ t \}) \cap X $ under $f_1$ for some $0 < c_2 \leq \min(c_1, \epsilon_2)$, since $f_1$ is continuous. Since $([0, c_2) \times (a_2, b_1)) \cap X$ is a union of connected line segments, where each connected line segment in this union contains a point in $([0, c_2) \times \{t \}) \cap X$, we find that $\tilde U_{2, i}$ also contains the image of $([0, c_2) \times (a_2, b_1)) \cap X$ under $f_1$.
We can construct a lift $g$ of $U_2$ to $\tilde X$ using the homeomorphism  $(p |_{\tilde U_{2, i}})^{-1} : U_2 \to \tilde U_{2, i}$. The restriction of $g$ to $\left([0, c_2) \times (a_2, b_2) \right) \cap X$) and the restriction of $f_1$ to $\left([0, c_2) \times [0, b_1) \right) \cap X$) agree on the intersection of their respective domains, since by the discussion in the previous paragraph, both of these lifts map $([0, c_2) \times (a_2, b_1)) \cap X$ into $\tilde U_{2, i}$. So they glue together continuously, giving us a lift $f_2$ that lifts $\left([0, c_2) \times [0, b_2) \right) \cap X$ homeomorphically into $\tilde X$.
$$ ... $$
Continuing in this way for $N$ steps, we end up with a lift $f_N$ on $\left([0, c_N) \times [0, 1] \right) \cap X$ for some $c_N > 0$, which maps this neighbourhood homeomorphically into $\tilde X$.

Finally, to show that $\tilde X$ is not simply connected, just pick any integer $k$ such that $\tfrac 1 k < c_N$. Consider the loop $\gamma$ in $X$ that is the rectangle
$$ (0, 0) \to (\tfrac 1 k, 0) \to (\tfrac 1 k, 1) \to (0, 1) \to (0, 0)$$
We can lift $\gamma$ to a loop $\tilde\gamma$ on $\tilde X$ using the lift $f_N$. Since the image of $\tilde\gamma$ under $p$ is $\gamma$, which is a non-trivial element of the fundamental group of $X$, $\tilde\gamma$ must be a non-trivial element of the fundamental group of $\tilde X$. This shows that $\tilde X$ is not simply connected.
