An extension of a question about no 3 points in $\mathbb{R}^2$ being collinear I am interested to see if we can use any of the answers here - or other methods - to answer the following question(s):
Does there exist a set $S\subset \mathbb{R}^2$ whose set of limit points in $\mathbb{R}^2$ contains the line $[0,1] =$ {$(x,0):  0 \le x \le 1$} such that no 3 points in S are collinear?
Update: I've deleted a silly follow-on question from the OP to simplify things for readers.
 A: To produce our set $S$, we "fly over" the interval $[0,1]$ again and again from left to right.  At the $n$-th stage, we add points with $x$-coordinates $\frac{k}{2^n}$, where $k$ ranges over the odd numbers from $0$ to $2^n$. The $y$-coordinates of the points added during the $n$-th stage are positive and have absolute value $\lt \frac{1}{2^n}$. 
At any time, we have only chosen a finite number of points, which determine a finite number of lines. So when we are choosing the $y$-coordinate of the point in $S$ corresponding to $x$-coordinate $\frac{k}{2^n}$, only finitely many points need to be avoided. 
Take any real $r$ in the interval $[0,1]$. For any $n$, there is a point of our set whose $x$-coordinate differs from $r$ by at most $\frac{1}{2^n}$, and therefore whose distance from the point $(r,0)$ is positive and $\lt \frac{\sqrt{2}}{2^n}$. So there is a sequence of points in $S$ with limit $(r,0)$.
A: The set $Q=\mathbb Q\cap(0,1)$ is a countable dense subset of $[0,1]$. Enumerate $Q=\{q_1,q_2,\ldots\}$, i.e. choose a bijection $q:\mathbb N\to Q$.  Now, let $$S=\{(q_n,\pi^{-n})|\;n\in \mathbb N\}.$$ No three points in $S$ are collinear, since this would mean $$(q_n,\pi^{-n})-(q_m,\pi^{-m})=\lambda((q_m,\pi^{-m})-(q_k,\pi^{-k}))$$ for some distinct natural numbers $n,m,k$ and $\lambda\in \mathbb R$. Comparing the first components, we get that $\lambda$ must be rational. Using this fact and comparing the second components, we obtain a nonzero polynomial with rational coefficients, that has $\pi$ as a zero. This contradicts the fact that $\pi$ is transcendental.
Closure of $S$ contains $[0,1]\times\{0\}$: choose an arbitrary open ball (in the $\infty$-norm for simplicity) centered at some $(x,0)$ where $x\in[0,1]$ with some (arbitrary) radius $r>0$. Then this ball must contain infinitely many points of $Q\times\{0\}$, since $Q$ is dense in $[0,1]$, so it contains $q_n$ with arbitrarily big $n$. Choose $n$ so that $\pi^{-n}<r$. Then $(q_n,\pi^{-n})$ lies inside the ball. This shows that for each $(x,0)\in[0,1]\times\{0\}$ there are points of $S$ arbitrarily close to $(x,0)$, thus proving the claim.
A: Yes, It is possible. Think of it like this, fisrt enumerate every rational in [0, 1] $q _ {0}, q_{1},q_{2}$... now proceed with consecutive choices, take $a_{0} = (q_0, 1)$ and $a_{n} = (q_n,y)$ whith y chosen so that it wont be collinear with any other couple of point and $|y|\le 2^{-n}$ it is alway possible becouse in the set $\{(q_0,t):0\lt t\le 2^{-n}\}$ only a finite number of points are subtracted for the non collinear request, this way for every (x,0) on your segment and for every $r \gt 0$ the ball centered on x whit radius r contain infinite point because it containt infinite rational point $(q_k, y_k)$ with k big enought so that $2^{-k} \lt r/4$ and $|x-q_k| \lt r/4$ thus in the ball.
actually you can manually choose the value of $y_k$, indeed you can take $y_k = sqrt(p_{k}/p_ {k}+1) $ this because all the $y_k$ are independent on R seen as vectorial space on Q; or also $y_k = \pi^{-k}$ for the same reason.
Note that every point with rational x-coordinate on the line generated by $(q_k,y_k)$ and $(q_j,y_j)$ are of the form $((1-a)*q_k + a*q_j),(1-a)*y_k + a*y_k)$ for some rational $a$.
