EDIT: I’ve got another framing that I thought would be more useful for later problems, but I was wrong. I still think there is some value in understanding this proof as well.
In particular, look at this diagram on Wikipedia. It would be better if the whole upper triangle was blue and the whole lower triangle were red instead of just one side (you can arbitrarily decide whether to paint the rest of the diagonal blue or red). If x=0 and x=1 aren’t fixed points, then they must be blue and red respectively. If we split [0,1] into n components of size 1/n, then we can see where f(x) maps each such component to and form a line of colored points as in q1. Proving this using Sperner’s Lemma is then essentially the same as qu2.
EDIT: I’ve got another framing that I thought would be more useful for later problems, but I was wrong. I still think there is some value in understanding this proof as well.
In particular, look at this diagram on Wikipedia. It would be better if the whole upper triangle was blue and the whole lower triangle were red instead of just one side (you can arbitrarily decide whether to paint the rest of the diagonal blue or red). If x=0 and x=1 aren’t fixed points, then they must be blue and red respectively. If we split [0,1] into n components of size 1/n, then we can see where f(x) maps each such component to and form a line of colored points as in q1. Proving this using Sperner’s Lemma is then essentially the same as qu2.