I'm trying to prove that $SO(n)$ is connected in the following way:
We'll approach by induction. For $n=1$, $SO(1)$ is just $\{\begin{bmatrix}1\end{bmatrix}\}$, so it's obviously connected. For $n=2$, $SO(2)$ is homeomorphic to $S^1$ (this can be verified by hand), and hence connected.
For $n>2$, let $A\in SO(n)$. If $e_1$ is an eigenvector of $A$, then the first column of $A$ is simply $e_1$. That is, $Ae_1=e_1$. Since $A$ is orthogonal, we conclude that $A^Te_1=A^{-1}e_1=A^{-1}Ae_1=e_1$ and hence $A$ is of the form $$A=\begin{bmatrix} 1&0\\ 0&B \end{bmatrix},$$ where $B$ is an element of $SO(n-1)$. Induction then yields the result.
Otherwise, consider the plane spanned by $e_1$ and $Ae_1$. Let $\{u_1,u_2\}$ be an orthonormal basis of this plane and let $R_1$ from the plane to itself denote the rotation taking $Ae_1$ to $e_1$.
Extend $\{u_1,u_2\}$ to an orthonormal basis $\{u_1,\dotsc,u_n\}$ of $\mathbb{R}^n$ and let $U$ be the matrix with columns $u_1,\dotsc,u_n$. Then, $$R=U\begin{bmatrix} R_1 & 0\\ 0 & I_{n-2} \end{bmatrix}U^T$$ is an element of $SO(n)$. Since $SO(2)$ is connected, there is an path from $R_1$ to $I_2$, which induces a path from $R$ to $I_n$ and thence from $A$ to $RA$.
To end the proof I just need to show that $$RA=\begin{bmatrix} 1&0\\ 0&B' \end{bmatrix},$$ where $B'$ is an element of $SO(n-1)$. In this case I could use induction to show a path from $RA$ to $I_n$.
However I can't seem to find why this is the case (that $RA$ has the desired form).
PS: I read all the MSE posts about the connectedness of $SO(n)$ and none of them solved my problem.