SO(n)/SO(n-1) is the sphere

Idea. $SO(n)$ acts transitively on the sphere with stabiliser of $e_1$ equal to $SO(n-1)$.

The map. Let $\varphi:SO(n)\to S^{n-1}$, $\varphi(U)=Ue_1$ (the first column). It is continuous and takes values in the unit sphere since $U$ is orthogonal.

Surjective. Let $v\in S^{n-1}$. Complete $v$ to an orthonormal basis $(v,w_2,\dots,w_n)$ and let $U_0$ be the matrix with these columns, $U_0\in O(n)$. If $\det U_0=-1$, replace $w_n$ by $-w_n$ (possible for $n\ge2$). Then $U_0\in SO(n)$ and $U_0e_1=v$.

The fibres are the cosets. Let $H=\{U\in SO(n):Ue_1=e_1\}$. For $U,V\in SO(n)$:
$$Ue_1=Ve_1\iff V^{-1}Ue_1=e_1\iff V^{-1}U\in H\iff U\in V\cdot H.$$
So $\varphi$ is constant on left cosets and its fibres are exactly the cosets: it induces a continuous bijection $\bar\varphi:SO(n)/H\to S^{n-1}$.

The stabiliser. If $Ue_1=e_1$ then (by orthogonality, $U^Te_1=e_1$ too) $U=\begin{pmatrix}1&0\\0&V\end{pmatrix}$ with $V\in O(n-1)$ and $\det V=\det U=1$. So $H\cong SO(n-1)$. For $n=2$, $H=\{I\}$ and the statement reads $SO(2)\cong S^1$.

Homeomorphism. $SO(n)$ is compact (closed and bounded), so $SO(n)/H$ is compact, $S^{n-1}$ is Hausdorff, and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Remark. $SO(3)/SO(2)\cong S^2$: choosing an axis (a point of the sphere) leaves a rotation about that axis undetermined.