Computing the fundamental group of the suspension of $X=\{0,1,1/2,1/3,1/4,...\}$. This is the problem 1.2.18 from Hatcher's algebraic topology.
Let $X=\{0,1,1/2,1/3,1/4,...\}$, inheriting a subspace topology from the real line, and $SX$ be its suspension, which looks like this:

We need to show that its fundamental group is free with a countable set of generators.
I tried Van Kampen theorem, but cannot find a good cover. Please give me some hints. Thank you.
 A: Let $X=\{x_0, x_1, \ldots\}$ be a countable discrete space.
Take $a_0 = [(x_0, 1)]\in SX$ to be a basepoint (the choice of $x_0$ doesn't matter, they all give the same point in $SX$).
Now for each $i\in\mathbb{N}$ define
$$V_i=\bigg\{(x_i, t)\ \bigg|\  \frac{1}{3} < t \leq 1\bigg\}$$
Note that each $V_i$ is open in $X\times[0,1]$. If we now go to the suspension then we can define
$$U_i=p\big(\{x_i\}\times[0,1]\big)\cup \bigcup_i p\big(V_i\big)$$
where $p:X\times[0,1]\to SX$ is the projection. So it's like a circle with  a countable umbrella at the top.
A little bit of gymnastics is needed to show that each $U_i$ is open in $SX$ and is homotopically equivalent to $S^1$. Also any interesection of the form $U_{i_1}\cap\cdots\cap U_{i_n}$ is contractible (as long as there are at least two different $U_i$ there). Every such intersection is an umbrella without full circle, you can contract it to the basepoint.
Thus van Kampen applies and therefore
$$\pi_1(X)\simeq *\pi_1(U_i)\simeq *_i\pi_1(S^1)\simeq *_i\mathbb{Z}$$
