Find seven points so any three contain two of distance one

Question

Choose 7 distinct points in the Euclidean plane so that among any 3 of those points, there are (at least) 2 that are a distance of exactly 1 apart.

Answer 1

First,

take two equilateral triangles with one side identified. This "diamond" gives an answer for four points with two "far" corners and two "near" corners

Then

consider two copies of this diamond in the plane. This has eight points but doesn't solve the puzzle because you could pick two far corners of one diamond and any point of the other diamond.

Next,

identify the far corner of one diamond with the far corner of the other diamond. Now we're good except if you take the identified point and the two other far corners.

Finally,

rotate one of the two diamonds until the other two far corners are the correct distance apart.

Explicitly,

assuming I did my arithmetic right, you could choose $A=(0,0)$, $B=(\sqrt{3}/2,1/2)$, $C=(\sqrt{3}/2,-1/2)$, $D=(\sqrt{3},0)$, $E=(\frac{5\sqrt{3}-\sqrt{11}}{12},\frac{\sqrt{33}+5}{12})$, $F=(\frac{5\sqrt{3}+\sqrt{11}}{12},\frac{\sqrt{33}-5}{12})$, and $G=(5\sqrt{3}/6,\sqrt{33}/6)$. Then the pairs $(A,B)$, $(A,C)$, $(B,C)$, $(B,D)$, $(C,D)$, $(A,E)$, $(A,F)$, $(E,F)$, $(E,G)$, $(F,G)$, and $(D,G)$ are all distance one apart.

@Quark suggested that I include an image (segments are of length one)

Answer 2

I agree with Gabriel's answer. I plotted the points just to verify them/explain to others visually; feel free to include the plot in your answer if you want and I'll delete this answer after. (red lines are lengths of 1)