Can it be derived from Hilbert's axioms that every plane contains a set of non-collinear points? 
*

*Can the following claim be derived from Hilbert's axioms as they are described in Wikipedia?

Every plane contains a set of non-collinear points.



*If the answer to the previous question is "no", can the following claim be derived from Hilbert's axioms?

In every plane that contains at least two distinct points $P$ and $Q$ there is a point non-collinear with $P$ and $Q$.


I posted a virtually identical question yesterday, but after a user pointed out in the comments that in Townsend's translation of Hilbert's axioms there is an axiom essentially stating the first claim, I concluded that the Wikipedia page was faulty, and deleted my question.
However after some research I've come to realize that Hilbert's axioms underwent refinements over the years, even after his death, and specifically Townsend's translation, from 1950, does not reflect the latest stage in this process of refinement.
The 13th edition of the German publication of Hilbert's axioms, from 1987 (this is the latest version I've been able to lay my eyes on), has no axiom that states that every plane contains a set of non-collinear points.
I believe the Wikipedia version of Hilbert's axioms reflects the latest "official" version of Hilbert's axioms, and, at any rate, it is with respect to this version that I request that my question be interpreted.
 A: Consider a plane $\alpha$.
Axiom $I.4$ tells us there exists some point $A$ lying on $\alpha$.
Axiom $I.3$ allows us to consider two distinct points $R_1, R_2$ which lie on some line. Clearly, both of them cannot be equal to $A$, since they are distinct; WLOG, let $R_1 \neq A$. Let $a$ be a line connecting $A$ and $R_1$, which exists by axiom $I.1$.
Axiom $I.3$ also allows us to consider three distinct points $Q_1, Q_2, Q_3$ which do not lie on a line. In particular, not all these points are on line $a$. WLOG, suppose $Q_1$ does not lie on $a$.
Axiom $I.4$ tells us there is some plane $\beta$ which contains $A, R_1, Q_1$. If $\beta = \alpha$, we are done; we have found three points $A, R_1, Q_1$ which do not lie on a line in plane $\alpha$.
Otherwise, we have $\beta \neq \alpha$. Then by axiom $I.7$, there is some other point $B$ which is on both $\alpha$ and $\beta$.
Now if both $R_1$ and $Q_1$ were in plane $\alpha$, we would have $\alpha = \beta$ by $I.5$. So one of those points is not in plane $\alpha$. That is, there is some point $C$ in $\beta$ but not in $\alpha$.
Axiom $I.8$ allows us to consider points $P_1, P_2, P_3, P_4$ which do not lie in a plane. In particular, they do not all lie in plane $\beta$. WLOG, suppose $P_1$ does not lie in plane $\beta$.
Suppose there were some line connecting $B, C, P_1$. We know that point $C$ is not in $\alpha$ while point $B$ is; therefore, $B \neq C$. Then by $I.6$, we would have that $P_1$ is in plane $\beta$. But this is a contradiction. So $B, C,$ and $P_1$ do not lie the same line.
Therefore, by $I.4$, there is some plane $\gamma$ on which $B, C$, and $P_1$ all lie.
Now since $C$ does not lie in $\alpha$ but does lie in $\gamma$, we see that $\gamma \neq \alpha$. But planes $\alpha$ and $\gamma$ do have a point in common; namely, $B$. Therefore, they have a second point $D \neq B$ in common.
Now suppose $A, B,$ and $D$ all lie on the same line. We know, therefore, that $A$ must lie on plane $\gamma$. But then both $\gamma$ and $\beta$ would include non-collinear points $A, B$, and $C$, and thus $\gamma = \beta$ by axiom $I.5$. We have already established this is not true.
So $A, B$, and $D$ are all points of $\alpha$ which are not collinear. The proof is complete. $\square$
