The chromatic number of any tree is 2
My proof for this is showing there is an algorithm to 2-color any graph. Equivalently, you can use induction to do the algorithm backwards.
Make a set (which starts empty) of "already colored points," and we'll call it C
Choose a point and put it in C with the first color.
While there are points not in C
Choose a point with an edge leading out of C
Color all vertices the point connects to with the color that the point is not
Put those points into C
End the While loop
The first thing that could go wrong is if there is nothing to choose. Luckily, this cannot go wrong because a tree is connected, and so there is a path between any two points of the graph. We know that C is not empty, and the while loop still goes while there are things not in C. And, there is a path between those two points, and there must be a switch from C to not C.
C is connected, from how it is constructed. Because the tree is connected, C ends when the points in C are the points in the tree, so it ends with a coloring.
It can only give each point one color, because at any point, a point not in C has at most one edge to C. If it had two, that would make a cycle, invalidating that we are working in a tree. So, any step of the algorithm has C being a valid coloring.
Induction:
A tree with one point clearly can be 2-colored.
If you assume any tree of size n-1 can be colored, then think of any tree of size n.
A tree always has a vertex of degree 1, so that vertex can be colored the other color than the one it is attached to.
So, a tree of size n is 2-colorable if a tree of size n-1 is 2-colorable, since one vertex is easy to color.
Combining a tree of size 1 is, and n-1 implies n, it is true for a tree of any size.