why it is a projective space? The unite sphere bundle of $TS^2$ , the tangent bundle of the 2-sphere, is the real projective space, $RP^3$. I can not understand the reason of it. And why the complement of the unit disc bundle of $TS^2$ is diffeomorphic to $(1,∞)×RP^3$
You should first convince yourself that the unit tangent bundle of $S^2$ is diffeomorphic to $SO(3)$. (Hint: There is a natural action of $SO(3)$ on the manifold of unit tangent vectors to $S^2$. What's the stabilizer subgroup of a point?)

Then it is a standard argument that $SO(3)\cong \Bbb RP^3$. Every nontrivial rotation has an axis and rotating $\pi$ around $\xi$ is the same as rotating $-\pi$ around $-\xi$. So $SO(3)$ can be identified with the unit ball $D^3$ with points on the boundary sphere identified to their antipodes. (Think "polar coordinates" with radius $\pi$.)
One places a plane separate to the sphere, and straight lines are formed by a projection of a feature when viewed from a point U.  Where U is interior to the sphere, the projection onto the map for a line running theough U gives a point on the map, and a pair of points or a through line in the sphere.

Where U is on the surface of the sphere, the projection gives the line at infinity for the plane tangent to the sphere at U, and a straight line equates to a circle through U.  The sphere is showing the inversion model, and the plane a general projective model.  When the tangent plame is parallel to the map, the model realy is euclidean geometry.

When U is outside the sphere, the results are the poincare model on the sphere, and the Beltrami Klein model on the map.  The cone with the apex at U and tangential to the sphere gives the hyperbolic Horocycle at infinity.  

You can see from this model, that it is possible to build a geometry, where it is not known the relation of U to the surface of the sphere, leaving it undetermined if real space is one of these three geometries