As the question says: I want to prove using Koenig's lemma that a rayless connected Graph $G$ with minimum degree $\delta(G) = k$ has a finite subgraph $H$ with $\delta(H) = k$.
So obviously this is trivial if $|G|$ was finite, so assume $|G|$ is infinite. Then I think I can use the contrapositive of Koenig's lemma somehow. The version of Koenig's lemma I'm familiar with is this: Let $V_0, V_1, V_2$ be an infinite sequence of disjoint non-empty finite sets, let $G$ be their union such that every $v$ in some $V_n$ has a neighbor $f(v)$ in $V_{n-1}$. Then $G$ contains a ray $v_0v_1v_2...$ with $v_n \in V_n$.
So here's my approach: I think the raylessness property means that for every given vertex $v \in G$ there are only finitely many "backward" paths of the form $xf(x)f(f(x))...v$ for otherwise there would be ray (One can construct one from infinitely many backward paths.)
But then I don't really know what I'm doing... Infinite combinatorics still confuses me quite a bit...