Reading the proof that $SO(3)\cong SU(2)/\{+I,-I\}$, I see that it hinges on the isomorphism theorem for groups. Since I am studying this in a Lie groups course, I would expect these two groups to be actually lie group isomorphic, and not just group isomorphic.
So my question is: is there any reason why the group isomorphism coming from the isomorphism theorem should be smooth?
It is clearly a Lie group homomorphism, that is, it is a differentiable map. You just consider the set $H=\{ai+bj+ck\,|\,a,b,c\in\mathbb{R}\}\subset\mathbb H$ and see $SU(2)$ as the quaternions with norm $1$. Now, each $q\in SU(2)$ acts on $H$ by $h\mapsto qhq^{-1}$ and it is easy to prove that the action of each $q$ is an element of $SO(3,\mathbb{R})$. This is cleary a differentiable map. Since $\{\pm\operatorname{Id}\}$ is a discrete subgroup of $SU(2)$, it induces a differentiable map from $SU(2)/\{\pm\operatorname{Id}\}$ into $SO(3,\mathbb{R})$. The fact that we use the isomorphism theorem to prove that this induces an isomorphism between $SU(2)/\{\pm\operatorname{Id}\}$ and $SO(3,\mathbb{R})$ changes nothing about that