For each n, if put it into one subsequence; if put it into the other subsequence. (You then have to prove that the first subsquence converges to 0, and the second subsequence converges to 1.)
Suppose {a_n} is a bounded sequence who's set of all subsequential
limits points is {0,1}. Prove that there exists two subsequences,
such that: one subsequence converges to 1 while the other converges
to 0, and each a_n belongs to exactly one of these subsequences.
Well, it's clear that at the limit points 0 and 1; there is a subsequence that that converges to it. Also, there is neighbourhoods of (-epps,epps) and (1-epps, 1+epps) around 0 and 1 respectively that contains infinitely many a_n's. I'm not quite sure about how to prove that each a_n belongs to exactly one of these subsequences or how to apply the bounded property of {a_n} into this question.