Finite Tree has Leaf Nodes

Theorem
Every non-edgeless finite tree has at least two leaf nodes, i.e., nodes of degree $1$.

Proof
We use the Method of Infinite Descent.

Suppose $T$ is a tree which has no nodes of degree $1$.

First note that no tree has all even nodes.

That is because by Characteristics of Eulerian Graph, it would then be an Eulerian graph, and by definition, trees do not have circuits.

From the Handshake Lemma, we know that $T$ must therefore have at least two odd nodes whose degree is at least $3$.

As $T$ is finite, this number must itself be finite.

Of those nodes, there must be two (call them $u$ and $v$) which can be joined by a path $P$ containing no odd nodes apart from $u$ and $v$.

(Otherwise you can pick as one of the two nodes one of those in the interior of $P$.)

Consider that path $P$ from $u$ to $v$.

As a tree has no circuits, all nodes of $P$ are distinct, or at least part of $P$ will describe a cycle.

Now consider the subgraph $S$ formed by removing the edges comprising $P$ from $T$, but leaving the nodes where they are.

The nodes $u$ and $v$ at either end of $P$ will no longer be odd, as they have both had one edge removed from them.

All the nodes on $P$ other than $u$ and $v$ will stay even.

The graph $S$ may become disconnected, and may even contain isolated nodes.

However, except for these isolated nodes (which became that way because of being nodes of degree $2$ on $P$), and however many components $S$ is now in, all the nodes of $S$ are still either even or odd with degree of $3$ or higher.

That is because by removing $P$, the only odd nodes we have affected are $u$ and $v$, which are now even.

Now, if the nodes in any component of $S$ are all even, that component is Eulerian.

Hence $S$ contains a circuit, and is therefore not a tree.

From Subgraph of Tree, it follows that $T$ can not be a tree after all.

However, if the nodes in any component $T'$ of $S$ are not all even, then there can't be as many odd nodes in it as there are in $T$ (because we have reduced the number by $2$).

Also, because of the method of construction of $T'$, all of its odd nodes are of degree of at least $3$.

By applying the Method of Infinite Descent, it follows that $T$ must contain a circuit, and is therefore not a tree.

So every tree must have at least one node of degree $1$.

Now, suppose that $T$ is a tree with exactly $1$ node of degree $1$. Call this node $u$.

From the Handshake Lemma, we know that $T$ must therefore have at least one odd node whose degree is at least $3$.

Let $P$ be a path from $u$ to any such odd node such that $P$ passes only through even nodes, as we did above.

Again, let us remove all the edges of $P$.

By a similar argument to the one above, we will once again have reduced the tree to one in which any remaining odd nodes all have degree of at least $3$.

Then we are in a position to apply the argument above.

Hence $T$ must have at least two nodes of degree $1$.

Note
It is instructive to see what happens to the above argument for a tree with exactly two nodes of degree $1$.

There may be no other odd nodes but these two (call them $u$ and $v$).

Such a graph is a path graph (which is itself a tree), and removing all the edges from the path from $u$ to $v$ leaves an edgeless graph, which of course has no circuits.

Alternative Proof
The theorem can also be proven by induction. Let the proposition we are proving in the case of a graph of order $k$ be denoted with $P\left({k}\right).$

Basis of the induction
$P\left({2}\right) $ is true since the tree of order 2 contains exactly 2 leaves (viz. its two nodes).

Induction hypothesis
$P\left({i}\right) $ is true $\forall i$ such that $2 \le i \le k$.

Induction step
Let $T_{k + 1}$ be a tree of order $k + 1$.

Pick two connected nodes of the tree, $u$ and $v$, and consider the subgraph $T_0$ obtained by removing the edge $e$ connecting $v_1$ and $v_2$.

By, Condition for Edge to be Bridge and the definition of tree, $e$ is a bridge and removing it results in a subgraph with two components.

By Subgraph of Tree each such component is a tree. Call $T_i$ the component containing $v_i$ and denote its order by $k_i$, where $i \in {1, 2}$

Since no nodes have been removed from $T_{k+1}$, $k_1 + k_2 = k + 1$.

There are two cases (and a trivial case):

Case 1
Neither $T_1$ nor $T_2$ is of order 1.

Then $2 \le k_i \le k - 1, i = 1, 2$.

But then by the induction hypothesis, both $T_1$ and $T_2$ have at least two leaves.

Either $v_1$ and $v_2$ are both leaves of the respective trees or one of them isn't.

Suppose that $v_1$ and $v_2$ are both leaves of the respective trees.

Then by adding $e$ back to the graph and thus recreating $T_{k + 1}$, $v_1$ and $v_2$ cease to be leaves.

However at least one other leaf from each subgraph $T_i, i = 1, 2$ survives the addition, becoming a leaf of $T_{k + 1}$.

Therefore $T_{k + 1}$ has at least two leaves.

Suppose that either $v_1$ or $v_2$ is not a leaf of its respective tree.

Then $T_{k + 1}$ has at least three leaves.

Case 2
$T_1$ or $T_2$ is of order 1.

Assume WLOG that $T_1$ is of order 1, i.e. $k_1 = 1$.

This implies that $k_2 = k$.

Then, by the induction hypothesis, $T_2$ has at least two leaf nodes.

Also $k_1 = 1$ implies that $v_1$ is a leaf of $T_{k + 1}$.

Suppose that $v_2$ is one of the leaves of $T_2$.

Then by adding $e$ back and thus recreating $T_{k + 1}$, $v_2$ is no longer a leaf, though at least one other leaf of $T_2$ survives the addition, becoming a leaf of $T_{k + 1}$.

Also, $v_1$ is again a leaf of $T_{k + 1}$.

Therefore $T_{k + 1}$ has at least two leaves.

Suppose that $v_2$ is not one of the leaves of $T_2$.

Then $T_{k + 1}$ has at least three leaves.

(The trivial case in which both $T_1$ and $T_2$ are of order 1 coincides with the basis of the induction).

In both non-trivial cases, the thesis follows by the Second Principle of Mathematical Induction.