Consider a sphere upon a plane (say $\mathbb{R}^2$). Let $C$ be the center of the sphere. Consider the lower hemisphere plus the boundary (bowl) and project lines from $C$ across the surface until they intersect the plane. We have a bijection between the bowl without border and $\mathbb{R}^2$. If we associate a direction to each pair of antipodal points on the boundary, we get a bijection between the bowl and $\mathbb{R}^2 \cup \{\mathrm{directions}\}$ (essentially, these are two models of the real projective plane)
Trying to make this bijection a homeomorphism, what kind of topology should be defined upon the bowl and $\mathbb{R}^2 \cup \{\mathrm{directions}\}$?
The slickest way to do this is simply to use quotient topologies.
On the bowl itself, use the quotient topology under which each border point of the bowl is identified with the antipodal border point. You obtain a quotient map $$q : \text{(bowl)} \to \text{(quotient space of the bowl)} $$ From this you also a bijection $$F : \text{(quotient space of the bowl)} \to \mathbb{R}^2 \cup \{\text{directions}\} $$ Using the bijection $F$, declare a subset $A \subset \mathbb{R}^2 \cup \{\text{directions}\}$ to be open if and only if it $F^{-1}(A)$ is an open subset of the quotient space of the bowl, if and only if $q^{-1}(F^{-1}(A))$ is an open subset of the bowl itself.
One thing you can observe is that $F \circ q$ is the same as the central projection that you describe, restricted to the bowl minus its border and with image being $\mathbb{R}^2$. It follows that if $A \subset \mathbb{R}^2$ then $A$ is an open subset of $\mathbb{R}^2$ if and only if $q^{-1}(F^{-1}(A))$ is an open subset of the bowl minus its border.
If this answer is a little too slick, it is also possible to describe more directly a basis for the topology on $\mathbb{R}^2 \cup \{\text{directions}\}$.