Number of pairs of points whose distance is one Let $S$ be a set of $n$ points in the plane, the distance between any two of which is at most one. Show that there are at most $n$ pairs of points of $S$ at distance exactly one.

My attempt on proof:
Construct a graph $G$ in which $2$ vertices are adjacent if and only if their distance is exactly one.
Then graph contains $3$ vertices in the form of equilateral triangle (so, $3$ pairs). I don't know how to generalize this.
 A: Hint 1:
given a point, where is located a point at distance 1 from this point ?
Hint 2:
considering the graph whose vertices are the points, and edges are the pairs of points at distance $1$, if a vertex has degree 3, show that one of its neighbours has degree one.
Hint 3:
once you have a cycle in this graph, can you have an other ?
[Edit, since a full proof was provided and mine was asked for]
I. intersection between unit circles.
If two distinct unit circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect, their intersection is made of two points, located on the bisector of the segment joining the centres of the circles. This intersection splits $\mathcal{C}_1$ into two arcs.The arc contained on the side of the bisector which does not contain the center of $\mathcal{C}_1$ is shorter than a semi-circle, and is contained in $\mathcal{C}_2$. The intersection of $k$ unit discs is a convex shape whose boundary is made of arcs from the $k$ circles. Each of these arcs is smaller than a semi-circle, and we will refer to such a shape as a polyarc. Since the discs we will intersect are distinct, there cannot be tangent intersections between their boundaries.
II. If a circle intersects an arc of a polyarc, a corner of the polyarc is outside the circle.
By convexity, the polyarc $\mathcal{P}$ and the circle $\mathcal{C}$ intersect in two points. If $\mathbf{p}_1$ is our intersection, by convexity as well, since all the corners of $\mathcal{P}$ are inside $\mathcal{C}$ the second intersection $\mathbf{p}_2$ is located on the same arc as $\mathbf{p}_1$. The portion of the arc between $\mathbf{p}_1$ and $\mathbf{p}_2$ is shorter than a semi circle, and therefore has to be the portion of the circle defining the arc which lies inside $\mathcal{C}$. Therefore, the two corners of the arc are outside $\mathcal{C}$, which is a contradiction.
III. If a vertex of the graph has valence 3, one of its neighbours has valence 1.
Let us consider $\mathbf{p}_0$ of valence 3, and $\mathbf{p}_1$, $\mathbf{p}_2$ and $\mathbf{p}_3$ its neighbours. Wlog, $\mathbf{p}_2$ is between $\mathbf{p}_1$ and $\mathbf{p}_3$ on the circle centered at $\mathbf{p}_0$. The polyarc resulting from the intersection of the discs of $\mathbf{p}_0$, $\mathbf{p}_1$ and $\mathbf{p}_3$ has 3 corners, and $\mathbf{p}_0$ is one of them. If $\mathbf{p}_2$ has a neighbour $\mathbf{p}_4$, the circle $\mathcal{C}$ centred at $\mathbf{p}_4$ passes through $\mathbf{p}_2$ and cuts an arc of the polyarc at this point. From II, one of the corners of the polyarc lies outside the circle, and this cannot be $\mathbf{p}_0$. Therefore half of the arc containing $\mathbf{p}_2$ is outside $\mathcal{C}$, and either does $\mathbf{p}_1$ or $\mathbf{p}_3$, which is a contradiction.
IV. There can be only one cycle
All the vertices of a cycle have valence at least 2, and are therefore corners of the polyarc. In addition, from II, adding a new vertex can only increase the number of corners of the polyarc by 1. Therefore the polyarc has no other corners than the vertices of the cycle. If a new vertex $\mathbf{p}_1$ is added strictly inside the polyarc and has a neighbour $\mathbf{p}_2$, the circle centered at $\mathbf{p}_2$ passing through $\mathbf{p}_1$ intersects the polyarc and a corner of the polyarc is outside this circle, which is a contradiction. $\mathbf{p}_1$ is therefore located on the boundary of the polyarc, and one of the corners is among its neighbours. Since the corners all have valence at least 2, $\mathbf{p}_1$ has valence 1 and cannot be part of a cycle. There is therefore only one cycle. If we have a cycle, cutting an edge of the cycle transforms the graph into a tree which has at $n-1$ vertices. Otherwise, the graph is a forest and has less than $n$ vertices.
A: While this seems to be a simple question, I do not know of an easy approach to it. This is the best that I could come up with.
Start off by constructing your graph with an edge between 2 vertices whenever their distance is 1.
Claim 1: If two edges have distinct endpoints, then these two edges must intersect.
Proof: Suppose not. Let the endpoints be $A, B, C, D$, where $AB$ and $CD$ are edges. Consider the convex hull of these points. If one of them is contained within (WLOG $A$), then we can show that $max(AD, AC) > AB $ which is a contradiction. Hence we must have 4 points on the convex hull, and labelled in clockwise order as (WLOG) $A, B, C, D$. Let $AC$ and $BD$ intersect at $X$. Then, by the triangle inequality, $AC + BD = XA + XC + XB + XC = (XA + XB) + (XC + XD) > 2$, which is a contradiction.
Claim 2: There are no even cycles (of length at least 4).
Proof: Suppose not. Let the even cycle be given by $v_1, v_2, \ldots v_{2n} $. Consider the line $v_1 v_2$. Then, by the previous claim, the odd and even vertices must be on different sides of $v_1 v_2$, as every consecutive pair must intersect this line. However, $v_1 v_{2n} $ and $v_2 v_3$ will not intersect, hence contradiction.
Claim 3: If there a cycle, then any edge that is not in this cycle must have an endpoint that is  contained within this cycle.
Proof: Suppose not, let the edge $AB$ have endpoints not contained in the cycle. By the first observation, each edge of the cycle must intersect $AB$, hence this splits the vertices must alternate above and below $AB$. This means that the cycle must have even length, which contradicts claim 2. 
Claim 4: We cannot have 2 different cycles. 
Proof: Suppose we do. Let the first cycle have vertices $v_1, v_2, \ldots v_{2n+1}$, and the second cycle have vertices $s_1, s_2, \ldots s_{2m+1} $. If any edge of the 2nd cycle has both endpoints in the first cycle, this will create an even cycle, which contradicts claim 2. But by claim 3, at least 1 endpoint of each edge must be in the first cycle. Hence, the endpoints of the 2nd cycle alternatively belong and don't belong to the first cycle. Hence, the 2nd cycle is even, which is a contradiction.
Claim 5: Any graph with at most 1 cycle has at most $n$ edges, where $n$ is the number of vertices.
Proof: If it has no cycle, it is a tree with at most $n-1$ edges.
If it has 1 cycle, remove 1 of the edges, and we get a tree with at most $n-1$ edges. Thus the original graph has at most $n$ edges.
Hence we are done.
A: Hint: To think about it another way, you're trying to prove that if there are $\gt n$ pairs of points of $S$ at distance exactly one, then there are at least two points that are greater than one unit away from each other.
