Claim: Given any chordal graph $G(V,E)$, the following algorithm will find a maximum cardinality set $S\subseteq V$ such that the induced subgraph of $S$ is $k$-colorable.
S <- emptyset
seq <- perfect elimination order of G
for x in seq:
S' <- S \cup {x}
if induced subgraph S' can be k-colored:
S <- S'
return S
Or in plain English: consider each vertex $x$ in perfect elimination order of $G$, every time adding $x$ to $S$ if possible.
The case of G being an interval graph has appeared on a competitive-programming problem (which means the above claim has been verified for interval graph case), however I didn't find any proof for its standard solution (I doubt there is any).
I've tried imitating proofs of similar algorithms on chordal graphs (namely algorithms for min coloring number, max clique and min clique cover), but without success.
I've also tried more "primitive" methods, among them is the following line of thinking: consider greedily picking vertices in perfect elimination order, all we need to prove is that the strategy of "keeping the earlier one whenever two vertices conflict (i.e. can't be both added into $S$)" is always the best strategy to resolve conflictions.
This approach also meets some difficulty: already-considered vertices can actually influence our current decision in complex ways, and I find it hard to deal with them.
Best Answer
For $k=1$, we are just constructing an independent set and everything should be fine, I think. For $k=2$, here is a counterexample:
The ordering $v_1, v_2, v_3, v_4, v_5$ is a perfect elimination ordering. Greedily, we would pick $\{v_1, v_2, v_3\}$, but then we can't pick $v_4$ or $v_5$. However, if we skip $v_3$, we can get the larger set $\{v_1, v_2, v_4, v_5\}$.