Why is spatial conformal infinity a point One property of spatial infinity is that all spacelike geodesics end at it. Since spacelike geodesics can have different directions, I do not understand why spatial infinity is a point. It looks more like a 2 sphere instead of a point. 

I will provide more information. Let us pick a point other than the spatial infinity in the conformal diagram. Usually, people draw the conformal diagram in a plane or represent it on the surface of a cylinder. So this point on a plane represents a 2 sphere. But spatial infinity is literally a point. Why? 
 A: Indeed the plane is conformal to the punctured sphere (by stereographic projection), rather than the open disc. This means that its conformal boundary is the single point at infinity on the sphere. This is an aspect of the uniformization theorem in 2-dimensions, but it's true in all dimensions.
To see why the plane is not conformal to the open disc, consider that a conformal map from the plane to the disc would be a bounded holomorphic function, and hence constant by Liouville's theorem in complex analysis.
In higher dimensions it follows another theorem of Liouville. Those hidden spheres of angles in the Penrose diagram you were asking about get squashed to zero size at infinity. Note that the situation is different for Minkowski space, whose conformal compactification has topology $S^1 \times S^d$. In general signature $(p,q)$ the compactification has topology $S^p \times S^q$. See this question for instance.
A: It may seem awkward or inconsistent that on a Penrose diagram,
$\mathscr{I}^+$ and $\mathscr{I}^-$ are shown as
lines (representing 3-dimensional things), while $i^0$, $i^+$, and $i^-$ are
 points (representing 2-spheres). The figure shows
why this actually makes sense.

Given a finite region
       of spacetime S, we can find a point like P that is spacelike with respect to the whole
       region, and a point like Q that is timelike with respect to the whole region. It is not possible to find a point that is lightlike in relation to every point is S.
This is obviously not a rigorous argument like Ryan Thorngren's, but hopefully it is helpful as a supplement to that answer, in order to build intuition. Please don't upvote this answer without upvoting his, since his is more rigorous and deserves the bounty.
A: Let me answer my question. 
By the definition of conformal flatness, $\nabla_a\Omega|_{i^0}=0$, where $\Omega$ is the conformal factor, and $i^0$ is the spatial infinity. So the spatial infinity is singular, and I think that is the reason people think spatial infinity is a point. 
