TITLE: Are all infinite graphs $3$-weak-edge colorable?
QUESTION [5 upvotes]: Let $G=(V,E)$ be a simple, undirected graph such that every vertex has degree at least $2$. Given $n\in\mathbb{N}$, a map $c:E \to \{1,\ldots, n\}$ is said to be a weak coloring if for every $v\in V$ the edges adjacent to $v$ do not all have the same color. (More formally, we want the restriction $c|_{E(v)}$ to be non-constant, where $E(v) = \{e\in E: v\in e\}$.)
These two nice posts by Mikail Tikhomirov and Brendan McKay respectively show that for every finite graph there is a weak edge coloring with $3$ colors. I tried to carry through their arguments with transfinite induction to infinite graphs - without success.
Question. If $G=(V,E)$ is an infinite simple undirected graph, is there  a weak edge coloring $c:E \to \{1,2,3\}$?

REPLY [2 votes]: Wlog $G$ is connected. Let $T\subseteq G$ be a spanning tree. Choose a vertex $r$ in $T$ with dgeree at least 2, make it the root. Split the neighbors of $r$ into two nonempty sets, $A$ and $B$. Color all tree edges between $r$ and $A$, blue, those between $r$ and $B$ red. Then continue coloring the edges of $T$: those from $A$ red, from $B$ blue, etc. Eventually all edges of $T$ are colored with red and blue. Finally color the remaining edges with green. 
This is good: all nonleaf vertices of $T$ have colors red and blue, the leaf vertices necessarily have a red/blue colored edge in $T$, and another in $G-T$, colored green.