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 $\chi(G) = \aleph\_0$?



Chromatic number of the linear graph on $[\omega]^\omega$
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. 


