I want to show that the fundamental group of SO(3) is $\mathbb{Z}/2\mathbb{Z}$ using the following theorem.
if $X$ is locally path connected and simply connected and $G$ is groups with a properly discontinous action over $X$ then $p:X\to X/G$ is a covering and $\Pi_1(X/G,x_0)=G$ where $x_0$ is a base point.
I think that I need to use $X=S^3$ but I don't understand how $SO(3)=S^3/(\mathbb{Z}/2\mathbb{Z})$. If it's not $S^3$ then which topological space should I use?
Ps: for properly distoninuos action I mean that $\forall x \in X\;\exists U(x)$ open set such that $g(U(x))\cap U(x)=\emptyset\;\forall g\neq e$ given $g\in G$
Take$$SU(2)=\left\{\begin{bmatrix}a&-\overline b\\b&\overline a\end{bmatrix}\,\middle|\,a,b\in\Bbb C\wedge|a|^2+|b|^2=1\right\}\simeq S^3.$$Ang let $G=\{\pm1\}\simeq\Bbb Z_2$. Then $G$ acts on $SU(2)$ in a natural way: $1$ acts as the identity and $-1$ acts is $q\mapsto-q$. This action is properly discontinuous. So, $\pi_1(S^3/\{\pm1\})\simeq\Bbb Z_2$.
Now, let$$H=\left\{\begin{bmatrix}\alpha i&-\beta+\gamma i\\\beta+\gamma i&-\alpha i\end{bmatrix}\,\middle|\,\alpha,\beta,\gamma\in\Bbb R\right\}.$$It turns out that if $q\in SU(2)$ and $h\in H$, then $qhq^{-1}\in H$. So, if $q\in SU(2)$, consider$$\begin{array}{rccc}r_q\colon&H&\longrightarrow&H\\&h&\mapsto&qhq^{-1}.\end{array}$$It turns out that, if $q,s\in SU(2)$, then $r_q=r_s\iff q=\pm s$. And, if$$\left\lVert\begin{bmatrix}\alpha i&-\beta+\gamma i\\\beta+\gamma i&-\alpha i\end{bmatrix}\right\rVert=\sqrt{\alpha^2+\beta^2+\gamma^2},$$each $r_q$ is an orthogonal map. So, you have a map from $SU(2)$ onto $SO(3,\Bbb R)$ which happens to be a covering map. So, $\pi_1\bigl(SO(3,\Bbb R)\bigr)\simeq\Bbb Z_2$.