2 clarified induction hypothesis

By AC, choose a cardinal well-ordering of the lines in in the plane and any well-ordering of all the points.

We proceed by transfinite induction.

Suppose $A_l$ is a set of points, no three colinear, and let $B_l$ be the set of lines spanned by points of $A_l$, and let $C_l=\cup B_l$. Suppose further that $l'\prec l$ implies $l'\in B_l$. Note that $|A_l| \leq |B_l| < |l|$, so that $$|C_l\cap l| = |\cup_{l'\in B_l} l\cap l'| < |l| .$$

• If $l\in B_l$, let $A_{Sl} = A_l$.
• If $a\in l \cap A_l$, take the minimal point $b\in l\backslash C_l$ and let $A_{Sl} = A_l\cup \{b\}$.
• If $A_l\cap l = \emptyset$, take the minimal two points $a,b\in l\backslash C_l$, and let $A_{Sl}=A_l\cup\{a,b\}$.
• Otherwise if $l'\$ is a limit ordinal, let $A_{l'} = \bigcup_{l\prec l'} A_l$.

It is easy to check that $A_{l'}$ has no three points colinear --- they'd have to all be in some $A_l$ for $l\prec l'$. The final union $\bigcup_l A_l$ has exactly two points on each line $l$.

1

By AC, choose a cardinal well-ordering of the lines in in the plane and any well-ordering of all the points.

We proceed by transfinite induction.

Suppose $A_l$ is a set of points, no three colinear, and let $B_l$ be the set of lines spanned by points of $A_l$, and let $C_l=\cup B_l$. Note that $|A_l| \leq |B_l| < |l|$, so that $$|C_l\cap l| = |\cup_{l'\in B_l} l\cap l'| < |l| .$$

• If $l\in B_l$, let $A_{Sl} = A_l$.
• If $a\in l \cap A_l$, take the minimal point $b\in l\backslash C_l$ and let $A_{Sl} = A_l\cup \{b\}$.
• If $A_l\cap l = \emptyset$, take the minimal two points $a,b\in l\backslash C_l$, and let $A_{Sl}=A_l\cup\{a,b\}$.
• Otherwise if $l'\$ is a limit ordinal, let $A_{l'} = \bigcup_{l\prec l'} A_l$.

It is easy to check that $A_{l'}$ has no three points colinear --- they'd have to all be in some $A_l$ for $l\prec l'$. The final union $\bigcup_l A_l$ has exactly two points on each line $l$.