Drawing planar graphs with integer edge lengths It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem).  Kemnitz and Harborth made the following stronger conjecture
Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.  
I was wondering if it is possible to attack this problem with the following approach.
Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear.  For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$


*

*$d(x_i, x_i') < \epsilon$,

*$d(x_i', x_j') \in \mathbb{Q}$, and

*no three points of $X'$ are colinear.


To prove Conjecture 2, it suffices to prove the following conjecture.
Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational.  Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.
Conjecture 3 is trivial for $n=1$ and easy for $n=2$.  Almering proved it for $n=3$, and I think it is open for $n>3$.  Note that Conjecture 3 is (essentially) a weakening of:
Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.
This question was posed by Ulam in 1945 (see this mathoverflow question for more background).  So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4.
Unfortunately, Conjecture 3 is beyond my limited area of expertise.  Hence:

Question. What are the prospects for proving Conjecture 3?  A proof or disproof would be fantastic.  However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.  

 A: Not quite an answer, but:


*

*The Kemnitz/Harborth conjecture was proved for cubic planar graphs in:


Straight line embeddings of cubic planar graphs with integer edge lengths
Jim Geelen1, Anjie Guo2,†, David McKinnon3
(Journal of Graph Theory, 2008)
They state a condition which would imply Kemnitz/Harborth (property 3.1 in their paper).
They cite the following theorem, which is related to, but not the same as, what you conjecture:
Theorem 2.1 (Berry 1992, Acta Arith). If $A, B, C \in \mathbb{R}^2$ are non-collinear points such that $dist(A, B), dist(A, C)^2,$ and $dist(B, C)^2$ are rational, then the set of points that are a rational distance from each of $A, B, C$ forms a dense subset of $\mathbb{R}^2.$
A: Problem D19 on pages 283-287 of Guy, Unsolved Problems In Number Theory, asks, "Is there a point all of whose distances from the corners of the unit square are rational?" This suggests that even a very weak form of Conjecture 3 is wide open (or was, as of 2004). 
A: I think conjecture 3 is actually stronger then conjecture 4. 
I prove $C_3\implies C_4$:
Pick any sequence of integers $a_n$, which contains all integers infinite times.
Pick any enumeration of all squares $s_n$ in the plane with corners at rational coordinates.
Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.
At step $\omega$, we have $\omega$ many rationals in every square, so a dense set of all rational distances.
