Is it possible to find an infinite set of points in the 3D space where the distance between any pair is rational? This is the question: Is it possible to find an infinite set of points in the 3D space, not all on the same plane, such that the distance between every pair of points is rational?
 A: Yes, such a set exists.
Take the infinite points on the circle of radius $1$ and centred at the origin in the plane $z=0$ given by lulu's answer in Is it possible to find an infinite set of points in the plane where the distance between any pair is rational?
Then add the point $(0,0,\sqrt{3})$.
So let
$$S=\{(\cos\theta,\sin\theta,0)\;:\; \theta\in [0,2\pi),\tan(\theta/4)\in\mathbb{Q}\}\cup
\{(0,0,\sqrt{3})\}$$
then
i) $S$ is infinite because $x\to\tan(x/4)$ is continuous and strictly increasing in $[0,2\pi)$.
ii) $S$ is non-planar because it contains $(0,0,\sqrt{3})$, $(1,0,0)$, $(-1,0,0)$, and $(\cos\theta_0,\sin\theta_0,0)$ with $\theta_0=\arctan(1/2)$.
iii) The distance between every pair of points of $S$ is rational. 
The distance between $(0,0,\sqrt{3})$ and $(\cos\theta,\sin\theta,0)$ is $\sqrt{\cos^2\theta+\sin^2\theta+3}=2$. Moreover the distance between $(\cos\alpha,\sin\alpha,0),(\cos\beta,\sin\beta,0)\in S$ is
$$\sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}=2\sqrt{\frac{1-\cos(\alpha-\beta)}{2}}\\
=2\left|\sin\left(\frac{\alpha-\beta}{2}\right)\right|
=\frac{4|\tan(\frac{\alpha}{4}-\frac{\beta}{4})|}{1+\tan^2(\frac{\alpha}{4}-\frac{\beta}{4})}\in\mathbb{Q}
$$
because 
$$\tan\left(\frac{\alpha}{4}-\frac{\beta}{4}\right)=\frac{\tan(\frac{\alpha}{4})-\tan(\frac{\beta}{4})}{1+\tan(\frac{\alpha}{4})\tan(\frac{\beta}{4})}\in\mathbb{Q}.$$
