I think I can grasp that the three axioms of a projective plane can be satisfied using this unit square: A bit of sketching and I can see that you can't draw two lines that do not intersect, and you can't draw two lines that intersect in more than one point.
I can also visualize the real projective plane as consisting of the positive component (say, along the $z$-axis) of all the lines that pass through the origin in $R^{3}$ (as that explains visual perspective--where vanishing point(s) are all on the line at infinity).
I've seen animations that transform the unit square to a hemisphere, and that seems to be relatable to the notion of the real projective plane as consisting of the positive component of all the lines that pass through the origin.
So one can obtain a hemisphere from both of these means of describing the real projective plane.
However, I'm a bit confused by the need to include a Mobius strip along the boundary of the hemisphere obtained from the unit square. I can see why it's necessary when I work with the unit square as a square, but I think I'm losing something when I visualize it being transformed into a hemisphere.
This is what I've come up with: If I picture the real projective plane as the surface of a hemisphere centered on the origin, with points consisting of the intersection of a line through the origin with the surface, then a point is replaced by its antipode whenever it crosses the equator of the hemisphere. The 'point' falls off the plane, so to speak, and is immediately replaced by its antipode on the opposite side of the hemisphere.
The Mobius strip, then, is the means of identifying opposite points on the equator of the hemisphere with each other.
Is this correct?