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.