This is Observation 3.23 from ''Canonical models for fragments of the Axiom of Choice'' from Paul Larson and Jindrich Zapletal.
We define the equivalence relation $=^+$ as follows: It connects $x,y\in X=(2^\omega)^\omega$ iff $rng(x)=rng(y)$. The graph $Z$ connects $x,y\in X$ if $\{x(n):n\in\omega\}=\{1-y(n):n\in\omega\}$ holds and $x=^+y$ fails.
Now let $\leq$ be a linear order on the $=^+$ quotient space. Define the function $c$ by letting $c(x)= \begin{cases} 0, & \text{for all } y\in X\text{ with }xZy\text{, }[y]_{=^+}<[x]_{=^+}\text{ holds,}\\ 1, & \text{else} \end{cases}$
for $x\in X$. Now it is claimed that it is not difficult to see that $c$ is a coloring of $Z$. Well but somehow I don't get to see it; so why is it so? Thanks for answering!
Suppose that $x\mathrel{Z}y$, if both have the same color, then it can't be $0$, so it must be $1$. Suppose also that $[y]_{=^+}<[x]_{=^+}$.
Now, $x$ itself is already a witness that $c(y)=1$, so there must be some $z$ such that $x\mathrel{Z}z$, but $[x]_{=^+}<[z]_{=^+}$.
But what does that mean?
Well, first of all it means that $\{1-y(n)\mid n\in\omega\}=\{x(n)\mid n\in\omega\}=\{1-z(n)\mid n\in\omega\}$. But that means that $y=^+z$, so $[y]_{=^+}=[z]_{=^+}$. But that's impossible, since this means $[y]_{=^+}<[x]_{=^+}<[z]_{=^+}=[y]_{=^+}$.