Is it possible to find an infinite set of points in the plane where the distance between any pair is rational?

You can even find infinitely many such points on the unit circle: Let $\mathscr S$ be the set of all points on the unit circle such that $\tan \left(\frac {\theta}4\right)\in \mathbb Q$. If $(\cos(\alpha),\sin(\alpha))$ and $(\cos(\beta),\sin(\beta))$ are two points on the circle then a little geometry tells us that the distance between them is (the absolute value of) $$2 \sin \left(\frac {\alpha}2\right)\cos \left(\frac {\beta}2\right)-2 \sin \left(\frac {\beta}2\right)\cos \left(\frac {\alpha}2\right)$$ and, if the points are both in $\mathscr S$ then this is rational.

Details: The distance formula is an immediate consequence of the fact that, if two points on the circle have an angle $\phi$ between them, then the distance between them is (the absolute value of) $2\sin \frac {\phi}2$. For the rationality note that $$z=\tan \frac {\phi}2 \implies \cos \phi= \frac {1-z^2}{1+z^2} \quad \& \quad \sin \phi= \frac {2z}{1+z^2}$$

Note: Of course $\mathscr S$ is dense on the circle. So far as I am aware, it is unknown whether you can find such a set which is dense on the entire plane.

Yes, it's possible. For instance, you could start with $(0,1)$ and $(0,0)$, and then put points along the $x$-axis, noting that there are infinitely many different right triangles with rational sides and one leg equal to $1$. For instance, $(3/4,0)$ will have distance $5/4$ to $(0,1)$.

This means that most if the points are on a single line (the $x$-axis), but one point, $(0,1)$, is not on that line.

There is essentially only one known infinite rational-distance set, built from rational Pythagorean triples, and all other examples are derived from this by inversions (with rational radius and center one of the points in the set), isometries, dilation, and taking subsets.

There are no other examples on algebraic curves (Solymosi and de Zeeuw 2008, http://arxiv.org/abs/0806.3095).