SO(n) is connected by Givens rotations

Notation. For $1\le i<j\le n$ and $\theta\in\mathbb{R}$, the *Givens rotation* $G_{ij}(\theta)$ is the identity matrix except for the four entries $(i,i),(i,j),(j,i),(j,j)$ which form the block $R(\theta)$. It lies in $SO(n)$ and $\theta\mapsto G_{ij}(\theta)$ is a path from $G_{ij}(0)=I$.

Induction. For $n=1$, $SO(1)=\{1\}$. Assume $SO(n-1)$ is path connected and let $U\in SO(n)$, $n\ge2$.

Step 1: rotate $v=Ue_1$ onto $e_1$. $v$ is a unit vector. For $j=n,n-1,\dots,2$ choose an angle $\theta_j$ such that the rotation $G_{j-1,j}(\theta_j)$ turns the vector obtained so far into a vector whose $j$-th coordinate is $0$: it acts only on coordinates $j-1$ and $j$, and a plane rotation can bring any vector $(x,y)$ to $(\sqrt{x^2+y^2},0)$. After these $n-1$ rotations, we get the vector $(r,0,\dots,0)$ with $r=\pm1$ (the norm is preserved), and for the last rotation we can choose the angle so that $r=+1$ (a rotation in the plane $(1,2)$ can send $(x,y)$ to $(\sqrt{x^2+y^2},0)$ and we are done). Thus $G\,Ue_1=e_1$ for $G=G_{12}(\theta_2)\cdots G_{n-1,n}(\theta_n)$.

Step 2: the remaining matrix. $GU\in SO(n)$ has first column $e_1$. Since its columns are orthonormal, its first row is $e_1^T$ too (the first row has norm $1$ and its first entry is $1$). So
$$GU=\begin{pmatrix}1&0\\0&U'\end{pmatrix},\quad U'\in O(n-1),\ \det U'=\det(GU)=1,$$
so $U'\in SO(n-1)$.

Step 3: paths. By induction there is a path $U'_t$ from $I_{n-1}$ to $U'$ in $SO(n-1)$, hence a path $1\oplus U'_t$ from $I$ to $GU$ in $SO(n)$. Also $G^{-1}=G_{n-1,n}(-\theta_n)\cdots G_{12}(-\theta_2)$ is joined to $I$ by the path $G_{n-1,n}(-s\theta_n)\cdots G_{12}(-s\theta_2)$, $s\in[0,1]$. Hence
$$s,t\longmapsto G(s)^{-1}\,(1\oplus U'_t)$$
(first run $t$, then $s$) is a path inside $SO(n)$ from $I$ to $G^{-1}(GU)=U$.

Conclusion. $SO(n)$ is path connected.

Remark. The proof is the same as the QR-decomposition by Givens rotations used in numerical linear algebra, and it shows that $SO(n)$ is generated by plane rotations.