Proof:

It is clear that $SO(2, \mathbb{R})$ is connected (why?). Turning to the case $n \geq 3$, we consider the action of $SO(n, \mathbb{R})$ on the standard unit sphere $S^{n-1}$ in $\mathbb{R}^n$ given by

$$(A, {\bf v}) \mapsto A{\bf v},$$

where $A \in SO(n, \mathbb{R})$ and ${\bf v} \in S^{n-1}$. It is an exercise for the student to check that this group action is transitive and that the stabilizer of the unit vector ${\bf\hat{e}}_n$ is the subgroup $K$ consisting of all those matrices in $SO(n, \mathbb{R})$ whose last column is ${\bf {\hat e}}_n$. The subgroup $K$ is homeomorphic to $SO(n-1, \mathbb{R})$ and so, by induction hypothesis, is connected. By exercise 3, the quotient space $SO(n, \mathbb{R})$ is homeomorphic to $S^{n-1}$ which is connected. So the theorem can be applied with $G = SO(n, \mathbb{R})$, $H = SO(n-1, \mathbb{R})$ and $G/H$ is the sphere $S^{n-1}$ with $n\geq 2$. $\square$