Let us say that an index i is bad, if every finite subsequence of s starting at i is red (i.e. for every j ≥ i we have χ(s_i ... s_j) = red). Two cases:
Case 1: there are infinitely many bad indices. Here we go to the first bad index then the second, and so on. The colour of w₀ does not matter, and since subsequent words start at a bad index, they will all be red.
Case 2: there are finitely many bad indices. Then there is some k which is larger than all bad indices. We start by going to k (again, the colour of w₀ does not matter). Since k is not bad, there must be some blue word starting at k. We take that one and move to a larger index. Again, that index is not bad. We repeat this process to find our sequence.
Nice proof, similar to the one I posted just now but simpler -- you realised that we only need one special category ("bad" rather than "hard-red" + "hard-blue"). Gonna leave mine up though :)
Case 1: there are infinitely many bad indices. Here we go to the first bad index then the second, and so on. The colour of w₀ does not matter, and since subsequent words start at a bad index, they will all be red.
Case 2: there are finitely many bad indices. Then there is some k which is larger than all bad indices. We start by going to k (again, the colour of w₀ does not matter). Since k is not bad, there must be some blue word starting at k. We take that one and move to a larger index. Again, that index is not bad. We repeat this process to find our sequence.