I am trying to get a better understanding of the homeomorphism of $SO(3)$ to the Real Projective Plane, so that ultimately I can show that $\pi_1(SO(3)) = \mathbb{Z}_2$.
From wikipedia and many other sources, $SO(3)$ is homeomorphic to $\mathbb{RP}^3$, but this section describes seemingly a topological space that is homeomorphic to $\mathbb{RP}^2$:
The Lie group SO(3) is diffeomorphic to the real projective space RP3.[3]
Consider the solid ball in R3 of radius π (that is, all points of R3 of distance π or less from the origin). Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball. Rotation through angles between 0 and −π correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through π and through −π are the same. So we identify (or "glue together") antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group.
Indeed, the ball with antipodal surface points identified is a smooth manifold, and this manifold is diffeomorphic to the rotation group.
The description above makes sense to me, each point in the the solid ball in $\mathbb{R}^3$, represents a rotation with the 'amount of rotation' given by it's distance from the centre and the line from the point to the centre identifying the axis of rotation. Then identifying the anti-podal points on the surface - this seems to be pretty much the exact construction of $\mathbb{RP}^2$! (Except the ball here has radius $\pi$).
So I guess my confusion is: Is $SO(3)$ homeomorphic to $\mathbb{RP}^2$ or $\mathbb{RP}^3$ or both?
To construct the projective plane, you take the sphere and identify antipodal points. That's not the same thing as what you're doing here, when you take the ball and identify antipodal points on the boundary sphere only. Then you get the interior of the the ball, which is homeomorphic to all of $\mathbb{R}^3$, together with a projective plane $\mathbb{RP}^2$ "at infinity", and this combined gives you projective space, $\mathbb{RP}^3$.
(By the way, $SO(3)$ is three-dimensional, so it would be strange if it were homeomorphic to a plane.)