Proof behind $S^n\cong SO(n+1)/SO(n)$

Solution 1:

It is basically the orbit–stabilizer theorem.

$SO(3)$ acts by rotations on $\mathbb R^3$. This action restricts to a transitive action on $S^2$. Fix a vector in $S^2$, say $e_1 = (1,0,0)$. One has a continuous map $SO(3) \to S^2$ given by $A \mapsto Ae_1$. The subgroup of $SO(3)$ stabilizing $e_1$, the "kernel" of this map, to abuse language, is the block-diagonal subgroup $H = \{1\} \times SO(2)$. It follows that the quotient $SO(3)/H$ is in continuous bijection with $S^2$. Because both spaces are compact Hausdorff, it is a homeomorphism.

As you seem to have noticed, there is nothing special about $n=3$ in this result.