Proof:

Once we show that $\psi: S^3 \longrightarrow SO(3, \mathbb{R})$ is surjective it follows from lemma (22.4) and the definition of real projective spaces that $SO(3, \mathbb{R})$ and $\mathbb{R}P^3$ are homeomorphic.

To prove the surjectivity of $\psi$, note that $S^3$ and $SO(3, \mathbb{R})$ are three dimensional manifolds and $\psi$ is a smooth map. We show that the derivative $D\psi(1)$ is an invertible linear map and so by the inverse function theorem the image must contain a neighborhood of the identity. We merely have to recall from lecture 5 that if a subgroup $H$ of a connected topological group $G$ contains a neighborhood of the identity then $H = G$.

We now turn to the proof that $D\psi(1)$ is a surjective linear transformation. We shall regard $\psi$ as a map from $\mathbb{R}^4$ to $SO(3, \mathbb{R}) \subset M(3, \mathbb{R})$ and compute its derivative at $1$. For a quaternion $h$ with sufficiently small norm,

$$\psi(1+h)v - \psi(1)v = \Vert 1+h\Vert^{-2}(v + hv + v{\overline h}) - v + O(\Vert h\Vert^2) = -2h_0v + hv + v{\overline h} + O(\Vert h\Vert^2),$$

where $h_0$ denotes the real part of $h$. We see that $D\psi(1)$ is the linear map $\mathbb{R}^4\longrightarrow M(3, \mathbb{R})$ given by

$$h\mapsto -2h_0(\cdot) + h(\cdot) + (\cdot){\overline h}. \eqno(22.4)$$

The kernel of this linear map contains $1$ and so is at-least one dimensional. It is exactly one dimensional since $D\psi(1)i$, $D\psi(1)j$ and $D\psi(1)k$ are linearly independent (skew-symmetric) matrices.