König's Tree Lemma/Proof 3

Proof
Suppose $T$ is a labeled tree with root $r$.

Let $L_0 = \left\{{r}\right\}, L_1, L_2, \ldots$ be the levels of $T$.

We construct our path by starting with $r$.

Next we partition all vertices in levels higher then $L_0$ (i.e. all vertices in levels $L_i$ where $i > 0$) into as many finite parts as is the cardinality of $L_1$.

In other words if $L_1 = \left\{{v_1, \ldots, v_n}\right\}$ then we partition the vertices (there are infinitely many of them) in $n$ parts:
 * $P_1, P_2, \ldots, P_n$

This partitioning is done as follows:

Any vertex $w$ will be connected to $r$ via some $v_i$ owing to connectedness.

We put $w$ in $P_i$.

Now since we have partitioned infinitely many vertices in finitely many parts, one of the parts will contain infinitely many vertices.

(This is also called the infinite pigeonhole principle).

Let that part be $P_j$.

Choose $w_1 = v_j$ as the next vertex in our under-construction path.

Next we delete the vertex $r$ and consider the component containing $w_1$.

It is also an infinite tree with each level finite and having $w_1$ as its root.

By a similar piece of reasoning, we get a vertex $w_2$ which is adjacent to $w_1$ and has infinitely many vertices corresponding to it.

We continue building our path by choosing $w_2$ as a part of it.

By induction we can continue and get an infinite path $r w_1 w_2 \cdots$.

This proves the theorem.