Configurations of eleven (or more) points in the Euclidean plane, such that out of any four there is a pair at unit distance. Inspired by this question, I was wondering the following:

What is the maximal size of a subset $C$ of the Euclidean plane such
  that out of any four points in $C$ there are two at unit distance from
  eachother?

I am able to show that for maximal $C$ we have $10\leq|C|\leq12$. The lower bound is easy:

The upper bound requires a lot more effort. I suspect the maximum is $11$, after spending a considerable amount of time fiddling around. Any ideas on how to approach this?
I'm also curious about replacing the requirement 'out of any four points' by 'out of any $n$ points', but first things first.
 A: I think your solution of 10 is optimal. This is not the solution, but I'm sketching a proof which I hope to finish soon.
Let's take a look at parts that the optimal solution might have. It goes without proof that each point is at least in a triangle in some optimal solution. It follows almost immediately that at least one quadrilateral consisting of two triangles that have two points in common is part of the solution (by adding one point to the triangle:
top
 /\
/  \
---- side
\  /
 \/  the quadrilateral, consisting of two triangles, needs to appear in the solution.

It is easy to see that the optimal solution does NOT have another point added to this quadrilateral at unit distance from 2 of these points. It would create this trapezium wherever you'd add such a point:
 /\
/  \
----\
\  / \
 \/___\ the trapezium cannot be part of the solution

The reason that this cannot be part of the solution, is that wherever we add yet another point, we then already have 3 points at non-unit distance from each other and it will bring us nowhere.
So some optimal solution consists only of triangles, has the quadrilateral but does not have the trapezium.
Now, let's add more triangles. They can have 0 or 1 point in common with the quadrilateral, but not 2.
First, let's check if two more triangles can both have the same point of the quadrilateral in common. However we do that, 3 points will be at non-unit distance from each other and it will lead nowhere (even if the two new triangles have another point in common with each other).
The same holds when adding two new triangles that each have another point of the quadrilateral in common.
Unfortunately, it also holds when adding two triangles that have one (in total) or no points in common with the quadrilateral, but at least we're then adding the maximum possible number of points for two extra triangles.
(TBC)
