I have thought a bit more about this since. Arrange all words in a rooted binary tree with the empty word Ɛ at the root and each vertex w having color χ(w) and children w0 and w1. Each cut [1] through the tree yields a set of words that can be used to cover all infinite binary sequences. Therefore, if there is a monochromatic cut, then we are done. To obstruct a monochromatic cut of color A, there must be a monochromatic obstructing path of color B from the root all the way down to infinity. To obstruct red and blue cuts, we need a red and a blue obstructing path, and you gave an example. So additional work is required to show that the task is also possible if the coloring contains a red and a blue cut-obstructing path.

[1] With cut I mean a set of vertices where no vertex in the cut is a descendant of any other vertex in the cut. Informally, repeatedly pick a vertex in the tree and remove the subtrees rooted at its children, the leaves of the remaining tree are the cut.