O(n) has exactly two components

Idea. The determinant splits $O(n)$ into two pieces; a reflection shows that the two pieces have the same shape.

Step 1: $O(n)$ is the disjoint union of two clopen sets. For $U\in O(n)$, $\det(U)^2=\det(U^TU)=1$, so $\det U=\pm1$. Since $\det$ is continuous, $SO(n)=\det^{-1}(1)$ and $O(n)\setminus SO(n)=\det^{-1}(-1)$ are closed and open in $O(n)$ and disjoint. The second is non-empty: it contains $D=\operatorname{diag}(-1,1,\dots,1)$. So $O(n)$ is disconnected.

Step 2: the second piece is connected. Consider $L:O(n)\to O(n)$, $L(U)=DU$ (note $D\in O(n)$, so the product stays in $O(n)$). Then $\det(DU)=-\det U$, so $L$ maps $SO(n)$ into $O(n)\setminus SO(n)$. It is onto that set: if $\det V=-1$ then $V=D(DV)$ with $\det(DV)=1$. So $O(n)\setminus SO(n)=L(SO(n))$ is the continuous image of a connected set, hence connected. (In fact $L$ is a homeomorphism, its inverse being $L$ itself.)

Step 3: conclusion. $O(n)$ is the disjoint union of two non-empty connected clopen sets, so these are exactly its two components.

Remark. The hypothesis is proved later (Problems 16 and 17). In the language of Lie groups, $SO(n)$ is the *identity component* of $O(n)$, and $O(n)/SO(n)\cong\mathbb{Z}/2$.