Let $[\omega]^\omega$ denote the set of infinite subsets of $\omega$. Let $$E = \{\{a,b\}: a,b\in [\omega]^\omega\text{ and } a\cap b = 1\}.$$ It is clear that $G = ([\omega]^\omega, E)$ has no uncountable cliques, but do we also have $\chi(G) = \aleph_0$?
1 Answer
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Yes.
Take the two smallest elements of a vertex $V\in[\omega]^\omega$ as its color. The number of colors is $\aleph_0$, and any two vertices with the same color shares at least two elements, so they are not connected.

$\begingroup$ Wonderfully short argument, thanks @bullet51! $\endgroup$ Commented Jun 12, 2019 at 11:34