Prove that every tree with $n\geq 2$ vertices has at least two vertices of degree one.
What I tried:
Suppose that there are fewer than two vertices of degree one. So we can split into two cases.
Case one: there is no vertex of degree one. Since we know that every tree has $n-1$ edges then the total degree of any tree have to be $2(n-1)$. But for this case since no vertex have degree $1$ then every vertex have at least a degree of $2$ and since there are $n$ vertices, the total degree is $\geq 2n$ which is a contradiction.
Case two: there is only one vertex of degree one. Similarly the total degree of any tree have to be $2(n-1)$. Then there are $(n-1)$ vertices with which have degree of $\geq 2$ while only one vertex with degree of one. Thus summing up to find the total of the vertices, we have that the total degree of vertices is $\geq 2(n-1)+1=2n-1$ which is also a contradiction.
This thus proves the statement for both cases.
Is my proof correct? Could anyone explain better?