T: Trees with no vertex of degree 2 have more leaves than internal nodes
So far I have (proof by contradiction).
Consider the opposite. That all nodes have only 1 leaf as a neighbor. Take some vertex, $v_i $. It must be connected to 2 other vertices which must be internal nodes (deg(3)). However this would mean there could exist a tree with more internal nodes (2) than leaves which would contradict T. Hence the original claim is false. Hence there is at least one vertex with 2 or 3 leaves.
Feel like this isn't as watertight as I would like.