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?
Yes. Normally I like to give some more motivation for an answer, but in this case it's probably hard to say anything suggestive that the illustration below does not, besides perhaps that after experimenting one might guess that the four-point "diamond" configurations are helpful (and probably necessary) in constructing such an arrangement.
(I remember seeing this problem, by the way, during a mail-in high school mathematics competition circa 1999; probably the widespread availability of the Internet makes it impractical to hold this sort of competition today.)
Edit I've posed a follow-up question, asking whether this is the unique configuration up to Euclidean motions. Servaes wrote an excellent, detailed answer showing that it is.
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)
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?