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 $l$.
The answer to your second question is yes. Let $G=(\mathbb{Q}[\sqrt{2}], +).$ This is a countable abelian group, so $\mathbb{R}/G$ is a hyperfinite Borel equivalence relation. In particular, it embeds into $\mathbb{R}/\mathbb{Q}.$ (See section 7 here for a summary of these results: Kechris - The theory of countable Borel equivalence relations, preliminary version May 8, 2019). Thus we can use a transversal for the latter to construct a transversal for the former. Let $X$ be a transversal for $\mathbb{R}/G.$
Let $V=\{(x + q\sqrt{2}) \text{ mod 1}: x \in X, q \in \mathbb{Q}\} \subset [0, 1).$ This is a Vitali set closed under translation by $\mathbb{Q}\sqrt{2}.$ We will show its outer measure is at least $\frac{1}{4}$ (in fact, the argument can be extended to show its outer measure is 1). Suppose towards contradiction that $U$ is an open cover of $V$ with measure less than $\frac{1}{4}.$ For each $n \ge 2,$ we can canonically find an open cover of $V \cap [0, \frac{\sqrt{2}}{n}]$ of measure less than $\frac{\sqrt{2}}{2n}$ by considering the least $m$ such that $\lambda(U \cap [\frac{m\sqrt{2}}{n},\frac{(m+1)\sqrt{2}}{n}])<\frac{\sqrt{2}}{2n}.$ With this we can recursively construct open covers $U_n$ of $V$ of measure less than $2^{-n-2}.$ Letting $\{q_n\}$ enumerate the rationals, we see $\{q_n+U_n\}$ covers $\mathbb{R},$ so $\lambda^*(\mathbb{R}) < 1,$ contradiction.
Also, here's an observation relevant to your first question. If $\mathbb{R}=\bigcup_{n<\omega} X_n$ is a countable union of countable sets, then there is a null subset of $[0, 1]$ which meets every mod $\mathbb{Q}$ class, namely $W=\bigcup_{n<\omega} \{x \in [0, 2^{-n}]: \exists q \in \mathbb{Q} (x+q \in X_n)\}.$ This is null since it countable outside of any $[0, \epsilon].$ Furthermore, if there is a Vitali set, that could be used to separate out a Vitali subset of $W,$ so a model satisfying these two properties would be a counterexample to the first question.
Best Answer
There is such a set which is Lebesgue measurable, and indeed of Lebesgue measure zero. To see this, start with a subset $S$ of $\mathbb R^2$ such that every line intersects it in continuum many points, for instance $C\times\mathbb R\cup\mathbb R\times C$, where $C$ is the Cantor set. Now repeat your favorite transfinite recursive construction of a set $A$ intersecting each line at exactly two points, modifying it in such a way that we all the points picked belong to $S$. Since $A$ is a subset of a measure zero set $S$, it itself is Lebesgue measurable of measure zero.
It appears that existence of such a set which is Borel is an open problem, see this MO post.