König's Lemma

Lemma
Let $$G$$ be a connected graph which is locally finite.

Then every vertex lies on a path of infinite length.

Proof
Let $$G$$ be a locally finite connected graph.

By definition, $$G$$ has an infinite number of vertices $$v_1, v_2, \ldots, v_k, \ldots$$, each of finite degree.

Let $$\mathcal{V}_k$$ be the set of all vertices adjacent to $$v_k$$.

As $$G$$ is a connected graph, between $$v_k$$ and every other vertex of $$G$$ there exists at least one path from $$v_k$$ to every other vertex of $$G$$.

Take any vertex of $$G$$ and call it $$v_1$$.

Let $$\mathcal{P}_1$$ be the set of all paths from $$v_1$$.

Each element of $$\mathcal{P}_1$$ must start with an edge joining $$v_1$$ to some element of $$\mathcal{V}_1$$.

There must be some $$v_r \in \mathcal{V}_1$$ such that there is an infinite path from $$v_r$$ in $$G$$ which does not pass through $$v_1$$. Otherwise, every path from $$v_1$$ would be finite, and since there is a path from $$v_1$$ to each other vertex of the graph, all vertices are contained within one of these finite paths. There are a finite number of paths from $$v_1$$, so all vertices of $$G$$ are contained within a finite set of finite sets, contradicting the assumption that $$G$$ is infinite.

By the axiom of dependent choice, we may pick one of the vertices of $$V_1$$ such that there exists an infinite path through it that does not include $$v_1$$, and call this $$v_2$$.

Each such infinite path must start with one of the elements of $$\mathcal{V}_2$$.

Repeating the above argument shows that there is some $$v_s \in \mathcal{V}_2$$ such that there is an infinite path from $$v_s$$ in $$G$$ which does not pass through $$v_2$$.

Thus we can construct by induction an infinite path.

The induction hypothesis states that there are infinitely many vertices which can be reached by a path from a particular vertex $$v_i$$ that does not go through one of a finite set of vertices.

The induction argument is that one of the vertices adjacent to $$v_i$$ satisfies the induction hypothesis, even when $$v_i$$ is added to the finite set.

The result of this induction argument is that for all $$n$$ we can choose a vertex $$v_n$$ as per the construction.

The set of vertices chosen in the construction is then a path, because each one was chosen to be adjacent to the previous one, and the construction guarantees that the same vertex is never chosen twice.

Note
If the graph $$G$$ is countably infinite, then the axiom of dependent choice is not required in order to be able to choose an infinite path from a given vertex that does not pass through a given finite set of vertices.