Locally Finite Connected Graph is Countable

Theorem
Let $G = \left({V, E}\right)$ be a graph which is connected and locally finite.

Then $G$ has countably many vertices and countably many edges.

Proof
We first show that $V$ is countable.

If $V$ is finite, then it is surely countable.

Suppose instead that $V$ is infinite.

Choose an arbitrary vertex $q \in V$.

Recursively define a sequence $\left\langle{S_n}\right\rangle$:

Let $S_0 = \{ q \}$.

Let $S_{n+1}$ be the set of all vertices that are adjacent to some element of $S_n$ but not adjacent to any element of $S_k$ for $k < n$.

That is, $S_n$ is the set of vertices whose shortest path(s) to $q$ have $n$ edges.

Since $G$ is connected, $V = \displaystyle \bigcup_{n \in \N} S_n$.

Furthermore, $S_n$ is finite and non-empty for each $n$.

But by Countable Union of Finite Sets is Countable, $\displaystyle \bigcup_{n \in \N} S_n$ is countable, so $V$ is countable.

By Cartesian Product of Countable Sets is Countable, $V \times V$ is countable.

Let $E' = \left\{{ (a, b): \left\{{a, b}\right\} \in E }\right\}$.

By Subset of Countable Set is Countable, $E'$ is countable.

Let $g: E' \to E$ be defined by $g(a, b) = \left\{{a, b}\right\}$.

Then $g$ is surjective.

Thus by Surjection from Natural Numbers iff Countable and Composite of Surjections is Surjection, $E$ is countable as well.