Seven points in the plane such that, among any three, two are a distance $1$ apart Is there a set of seven points in the plane such that, among any three of these points, there are two, $P, R$, which are distance $1$ apart?

EDIT: This answer (which starts with a bold "No.") is wrong as can be seen from Travis's answer (which starts with a justified "Yes."). In principle, I should delete it, but - for the moment - I'll leave it here because somebody may find it helpful in other ways, such as seeing how to arrive at some neccessary conditions (cf. some comments)

No.
Assume $d(A,B)$, $d(A,C)$, $d(A,D)$, $d(A,E)$ are all $\ne 1$. Then all distances amoing $B,C,D,E$ must be $=1$.
However, because the tetrahedron cannod be "flattened",

Among any four points there must be two of distance $\ne 1$.

We conclude that $A$ can have at most three points at distance $\ne 1$, hence 

Every point hast at least three points at distance $=1$.

Hence if we pick any two points at distance $\ne1$, then one of the remaining five points must be one of the intersection points of the $1$-circles around them, i.e., 

For any two points $A,B$ with $d(A,B)\ne 1$ there exists $C$ with $d(A,C)=d(B,C)=1$.

Can you take it from here?
