Proof:

Since $S^3$ is compact and connected, the image of the map $\psi:S^3 \longrightarrow O(3, \mathbb{R})$ is a compact and connected subgroup of $O(3, \mathbb{R})$. Now $O(3, \mathbb{R})$ is disconnected with two components and so the image must lie entirely in one of these components. Since $\psi(1)$ is the identity map this connected subgroup meets $SO(3, \mathbb{R})$ and so must be contained entirely in $SO(3, \mathbb{R})$.

Slightly more difficult is the proof that the image of $S^3$ under $\psi$ is the whole of $SO(3, \mathbb{R})$. It is possible to give an argument which uses only linear algebra but we prefer to follow a slightly more sophisticated approach using the inverse function theorem. The student who is uncomfortable may merely skim through the argument and take the result on faith.