Let T be a tree with 2k leaves. Prove that there exists k pairwise edge-disjoint paths in T connecting the leaves in pairs.
My attempt: Assume by contrary, let $p_i$ and $p_j$ be paths having an edge in common for $i\neq j$. Then the paths are overlapping as they determine 4 distinct leaves. Hence we proved the statement. Am I correct guys or I need to modify my arguments.
You can do this by induction on $k$. I outline the proof leaving a couple of steps for you to do.
If $k=0$, the tree has no leaves, so it is the empty tree, and there are obviously no path joining any leaves.
Assume that the theorem is true of a tree with $2k$ leaves, and that we are given a tree with $2(k+1)$ leaves. Recall that the depth of a node is the length of the shortest path from the node to a leaf, so that a leaf has depth $0$, the parent of a leaf has depth $1$ and so on. There must be a node with more than one child. (Prove this!) Choose a such a node $v$ of minimum depth. If we eliminate two of the paths from $v$ down to the leaves, say to $l_1$ and $l_2$, we are left with a tree with $2k$ leaves, which can be partitioned into $k$ edge-disjoint paths joining leaves, by the induction hypothesis. These paths are all edge disjoint from the path joining $l_1$ and $l_2$. (Prove!) This gives $k+1$ edge-disjoint paths in all, proving the theorem.
If you don't like the idea of an empty tree, you can start with a tree with two leaves for the basis. This is proved just like the general case, starting with a node of minimum depth with two children.