One can use the double covering of $\text {SO}(3)$ by $\text {SU}(2)$ to compute the fundamental group of $\text {SO}(3)$.
I’d like to travel in the opposite direction, using the definition of $\text {SO}(3)$ to compute its fundamental group and show that the universal cover is isomorphic to $\text {SU}(2)$.
I want an approach that does not pull the $\text {SO}(3)$ to $\text {SU}(2)$ relationship or its equivalent in terms of quaternions or Clifford algebras out of thin air and then shows that everthing works.
It is standard that $SO(3)$ is homeomorphic to $\mathbb{R}P^3$. The abelianization of the fundamental group is thus $H_1(\mathbb{R}P^3)=\mathbb{Z}/2$. Since these are topological groups, the fundamental groups are already abelian, so $\pi_1(SO(3))=\mathbb{Z}/2$.
Or if you are okay with using the fact $S^2$ covers $\mathbb{R}P^2$, then you can use the fact $\mathbb{R}P^3$ has the same $2$-skeleton as $\mathbb{R}P^2$.