Proof

We give a proof by contradiction. Let \(T\) be a tree with at least two vertices, and suppose, contrary to stipulation, that there are not two vertices of degree one.

Let \(P\) be a path in \(T\) of longest possible length. Let \(u\) and \(v\) be the endpoints of the path. Since \(T\) does not have two vertices of degree one, at least one of these must have degree two or higher. Say that it is \(u\text{.}\) We know that \(u\) is adjacent to a vertex in the path \(P\text{,}\) but now it must also be adjacent to another vertex, call it \(u'\text{.}\)

Where is \(u'\text{?}\) It cannot be a vertex of \(P\text{,}\) because if it was, there would be two distinct paths from \(u\) to \(u'\text{:}\) the edge between them, and the first part of \(P\) (up to \(u'\)). But \(u'\) also cannot be outside of \(P\text{,}\) for if it was, there would be a path from \(u'\) to \(v\) that was longer than \(P\text{,}\) which has longest possible length.

This contradiction proves that there must be at least two vertices of degree one. In fact, we can say a little more: \(u\) and \(v\) must both have degree one.