I'm trying to prove that the two subsets defined by the plane separation axiom are not empty.
In order to be in the same page, the axiom that I have says the following:
For every $r\subset P$ (where $r$ is a line and $P$ is the plane), there are two subsets $H_{1}$ and $H_{2}$ that have the following properties:
$(1) H_{1}\cup H_{2}=P-r$
$(2) X,Y\in H_{i} \implies$ [X,Y] $\subset H_{i}$
$(3) X \in H_{1}, Y \in H_{2} \implies$ [X,Y]$\cap r\ne \emptyset $
Now, my reasoning goes as follow. Since the plane has at least three different points that are not aligned, $P-r\ne\emptyset$. Hence, at least one of the two sets, e.g. $H_{1}$, is not empty. Taking $A \in H_{1}$ and $B \in r$, we can draw the line $r_{AB}$ on which we can find a $C$ point that verifies that $B$ is the middle point of [A,C].
So we found a $C\in P-r$ such that [A,C] cuts $r$.
My question is, how we can conclude from here that $C \in H_{2}$? It seems pretty obvius but I don't know how to put it formally.
Thank you