Here is a different way to carry out this argument. Consider equations (3) and (4) together.

Eq. (4) says that $\delta q$ is orthogonal to $q$; indeed, this makes intuitive sense: $q$ is constrained to have unit length and so lies on the surface of the hyper-sphere. If $q$ is on its surface already, then only the infinitesimal movements in the orthogonal directions will remain on the surface.

Eq. (3) says that $\delta q$ is orthogonal to $Rq$. Note that infinitesimal vector $\delta q$ can be arbitrary, as long as satisfies the above constraint of being orthogonal to $q$. For any such $\delta q$ it must be that it is also orthogonal to $Rq$. The one and only way this can be true, is if $Rq$ is parallel to $q$.

Which makes $q$ an eigenvector of $R$ by definition.

Here is a slightly different way to carry out this argument. Consider equations (3) and (4) together: $$\delta q ^\top R q = 0 \\ \delta q ^\top q = 0$$

Eq. (4) says that $\delta q$ is orthogonal to $q$; indeed, this makes intuitive sense: $q$ is constrained to have unit length and so lies on the surface of the hyper-sphere. If $q$ is on its surface already, then only the infinitesimal movements in the orthogonal directions will remain on the surface.

Eq. (3) says that $\delta q$ is orthogonal to $Rq$. Note that infinitesimal vector $\delta q$ can be arbitrary, as long as satisfies the above constraint of being orthogonal to $q$. For any such $\delta q$ it must be that it is also orthogonal to $Rq$. The one and only way this can be true, is if $Rq$ is parallel to $q$.

Which makes $q$ an eigenvector of $R$ by definition.