Suppose (xn) is a bounded sequence that does not converge. Show that (xn) has at least two subsequences that converge to different limits. By the Bolzano-Weierstrass theorem there exists a subsequence of (xn) that converges. Delete that subsequence from (xn) and form a new sequence (xn1) with the remaining terms. (xn1) is bounded and hence has a convergent subsequence. If that subsequence converges to a different limit then we are done. If it does not, delete that subsequence from (xn1) and form a new sequence (xn2) with the remaining terms; ad infinitum. My argument is that not all of those subsequences can converge to the same limit, because the main sequence (xn) is not convergent. I dug up this problem while flipping through my old analysis book. What did I do wrong? I feel as though something is not right with this proof. I found another proof: There exists a convergent subsequence by B-W, converging to some number x. Since (xn) does not converge, it has a tail that is at least e>0 away from x. From that tail there exists another convergent subsequence which must be at least e away from x, hence there are two convergent subsequences.