Alternative picture
This illustrates the fact that $\mathbb{R}P^1$ is homeomorphic to $S^1$. Here we regard $S^1$ as the space of complex numbers $z$ with $|z|=1$. Recall that $\mathbb{R}P^1=S^1/E$, where $uEv$ iff $v=\pm u$. A point of $\mathbb{R}P^1$ is an equivalence class for this relation, or in other words a pair of opposite points in $S^1$, as shown on the left. Given a single point $z\in S^1$, as shown on the right, the two square roots of $z$ form an equivalence class, as shown on the left. By dragging the right hand point you can see how the left hand pair varies continuously.