SO(n) is path connected (normal form)

Idea. Any rotation is a product of commuting rotations in orthogonal planes, and each can be "turned back" to the identity continuously.

Step 1: normal form. Let $U\in O(n)$. Then there are $Q\in O(n)$ and angles $\theta_1,\dots,\theta_p\in(0,\pi)$ such that
$$U=Q\,\operatorname{diag}\big(R(\theta_1),\dots,R(\theta_p),\,I_a,\,-I_b\big)\,Q^T,\qquad R(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix},$$
with $2p+a+b=n$. (Derivation: $U$ is a normal matrix, so it is unitarily diagonalisable over $\mathbb{C}$ with eigenvalues of modulus $1$. Real eigenvalues are $\pm1$ and non-real ones come in conjugate pairs $e^{\pm i\theta}$. For an eigenvector $v=x+iy$ of $e^{i\theta}$, the real vectors $x,y$ span a $U$-invariant plane on which $U$ acts as a rotation by $\theta$, in an orthonormal basis of that plane. Eigenspaces for distinct eigenvalues are orthogonal, which gives an orthonormal basis of $\mathbb{R}^n$.)

Step 2: for $U\in SO(n)$, the multiplicity $b$ of $-1$ is even. $\det U=\det R(\theta_1)\cdots\det R(\theta_p)\cdot1^a(-1)^b=(-1)^b$, because each $\det R(\theta_j)=1$. So $\det U=1$ forces $b$ even, say $b=2m$.

Step 3: write $-I_{2m}$ as rotations. $-I_2=R(\pi)$, so $-I_{2m}=\operatorname{diag}(R(\pi),\dots,R(\pi))$. We obtain $U=Q\,B\,Q^T$ with
$$B=\operatorname{diag}\big(R(\theta_1),\dots,R(\theta_p),R(\pi),\dots,R(\pi),I_a\big)=\operatorname{diag}\big(R(\varphi_1),\dots,R(\varphi_q),I_a\big),\quad q=p+m.$$

Step 4: the path. Let $B_t=\operatorname{diag}\big(R(t\varphi_1),\dots,R(t\varphi_q),I_a\big)$, $t\in[0,1]$. It is continuous in $t$, $B_t$ is orthogonal with $\det B_t=1$, $B_0=I$ and $B_1=B$. Then
$$\gamma(t)=QB_tQ^T$$
is a continuous path from $I$ to $U$. It stays in $SO(n)$: $\gamma(t)$ is orthogonal and $\det\gamma(t)=\det B_t=1$. (Note that $Q\in O(n)$ may have determinant $-1$; conjugation does not change the determinant.)

Conclusion. Every element of $SO(n)$ is joined to $I$ in $SO(n)$, so $SO(n)$ is path connected.

Remark. The proof also shows that $\exp:\mathfrak{so}(n)\to SO(n)$ is onto: $\gamma(t)=\exp(tX)$ with $X=Q\operatorname{diag}(\varphi_1J,\dots,\varphi_qJ,0)Q^T$, $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ (Problem 8 of the exponential chapter). Compare with $SO(n)$ being compact and connected.