Cardinality of Sets of Points in Neutral Geometry I recently proved the following result:
(1) Let $M$ be a model of Neutral (Absolute) Geometry with set of points $\mathbb{P}$. Then $|\mathbb{P}|= \mathfrak{c}$.
However, the proof relies upon the following theorem:
(2) If $κ$ and $λ$ are cardinal numbers such that $κ$ is infinite and $0<λ≤κ$, the union of $λ$ sets of cardinality $κ$ has cardinality $κ$. In other words, the union of at least one and at most $κ$ sets of cardinality $κ$ has cardinality $κ$ (First answer here).
My question is this: I don't know which axioms are necessary to prove (2) but I'm certain they aren't precisely those of Neutral Geometry. Assuming my proof contains no errors, is (1) still a valid theorem? Loosely speaking, I think the answer is yes since (2) was, in the proof, never used to draw any geometric conclusions but I'm not sure.
EDIT: The axioms of Neutral Geometry can be found here
The following terms are undefined: Distance, point, line, half-plane, angle measure and area

This answer refers to Hilbert axioms of neutral geometry, however maybe your theory is equivalent to Hilbert's.
It can be proved that neutral geometry has exactly two models up to isomorphism. Adding Euclid's fifth postulate or its negation we get a theory which is categorical, which means every two models are isomorphic. $\mathbb{R}^2$ and Klein model are the models of euclidian and hyperbolic geometry respectively and both of them are of cardinality $\mathfrak{c}$.
However you don't really need to know that this theory is categorical to prove that the set $P$ of all points has the cardinality $\mathfrak{c}$. All you need is some theorems which can be derived from the axioms. Take a line $L$ and a point $o\in L$. Let $A$ be one of the halflines contained in $L$ beginning in $o$ and $M$ be one of the halfplanes determined by $L$. For a given point $x\in P$ let $p(x)$ be an ortogonal projection on a line $L$ and define $$f(x)=\begin{cases}|xp(x)| & , x\in M \\ 0 & , x\in L \\ -|ap(x)| & , x\in M^*\end{cases}$$
$$g(x)=\begin{cases}|op(x)| & , p(x)\in A \\ 0 & , p(x)=o \\ -|op(x)| & , p(x)\in A^*\end{cases}$$
Then $x\mapsto (f(x),g(x))$ is a bijection between $P$ and $\mathbb{R}^2$.
