SO(2) is a circle, O(2) has two components

Step 1: $SO(2)=\{R_\theta\}$. Let $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SO(2)$. The first column $(a,c)^T$ is a unit vector, so $(a,c)=(\cos\theta,\sin\theta)$ for some $\theta$. The second column is a unit vector orthogonal to the first, so it is $\pm(-\sin\theta,\cos\theta)$, and $\det U=ad-bc=1$ selects the sign $+$. So $U=R_\theta$. Conversely $R_\theta^TR_\theta=I$ and $\det R_\theta=\cos^2\theta+\sin^2\theta=1$.

Step 2: $SO(2)$ is path connected. $\theta\mapsto R_\theta$ is continuous from $[0,2\pi]$ onto $SO(2)$, and the continuous image of an interval is path connected.

Step 3: the other half of $O(2)$. Let $S=\operatorname{diag}(1,-1)$. If $V\in O(2)$ with $\det V=-1$, then $\det(VS)=1$ so $VS=R_\theta$ for some $\theta$, and $V=R_\theta S^{-1}=R_\theta S$. Hence
$$O(2)\setminus SO(2)=\{R_\theta S:\theta\in\mathbb{R}\},$$
the image of $[0,2\pi]$ under the continuous map $\theta\mapsto R_\theta S$, hence path connected.

Step 4: exactly two components. $\det:O(2)\to\{\pm1\}$ is continuous. The two pieces $SO(2)=\det^{-1}(1)$ and $O(2)\setminus SO(2)=\det^{-1}(-1)$ are disjoint, non-empty, and both open and closed in $O(2)$. Each is path connected, so they are the two path components.

Remark. The map $\theta\mapsto R_\theta$ identifies $SO(2)$ with the circle $\mathbb{R}/2\pi\mathbb{Z}$, a *compact* connected group. This is the basic example for the study of $SO(n)$ in the following problems.