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.
Best Answer
This is the subject well-known open problem called the Hadwiger-Nelson Problem. The problem asks for the exact minimum number of colors that we can color the plane with, so that no two points of distance one are the same color.
You have observed that we cannot do it with only 2 colors, and asked, can we do it with 3 colors? The answer to that is no as well: on the Wikipedia page, we see that this is proven with the Mosser Spindle. It is a collection of 7 points in the plane, with 11 edges of length 1 between them, such that these points cannot be colored with only 3 colors.
Therefore it is known that 2 or 3 colors are impossible. It is not known, however, if it can be done with 4 colors. It is known that it can be done in 7 colors, so the minimum number of colors is either 4, 5, 6, or 7 -- but we don't know which!
In fact, it is expected by some that the exact minimum depends on the infamous Axiom of Choice.
UPDATE: Remarkable news earlier this year in April 2018: amateur mathematician Aubrey de Grey showed that 4 colors is impossible, thus narrowing the minimum down to 5, 6, or 7. To do this, he constructed a set of about 1500 points in the plane and proved that it is impossible to color them with 4 colors. This is the first progress on the problem in 60 years. You can read about it in de Grey's paper (which is quite readable). You can also read about it in this blog post and this Quanta magazine article.