I have the following claim which I have been struggling to write a proof for:
Let $G=(V,E)$ be a graph such that $deg(v)<\infty$ for all $v\in V$. Then every connected component in the graph is countable.
So my reasoning was along the line of choosing a general vertex $v\in V$ and showing that its connected component $[v]_G$ is countable by writing it as $[v]_G=\sqcup_{n=0}^\infty D_n(v)$, where $D_n(v)$ is the set of vertices in $G$ at distance $n$ from $v$. Since $D_n(v)$ is finite at all, $[v]_G$ is a countable union of finite sets and hence countable.
My issue is with writing why $D_n(v)$ is finite rigorously. I would like to use a straightforward argument of giving an upper bound which is countable, but since the degree is only finite and not bounded I think to say this one has to use an induction argument.
I would appreciate if any one has some direct, non inductive, argument to show that $D_n(v)$ is finite. If any one thinks there is an error in any of the above arguments, I would also be thankful if someone were to point them out.