Q: Proof that a convergent sequence has a monotones subsequence? I want to prove that every convergent sequence has a monotone subsequence. My idea is to trap some terms of the convergent sequence in successively smaller intervals around the limit.
Let $\lim_{n\to\infty} a_n=A$, and suppose that $(a_n)$ is not eventually constant. There must be infinitely terms of $(a_n)$ in at least one of the intervals $[A-1, A)$, $(A, A+1]$. 
Suppose there are infinitely many terms in $(A, A+1]$. Let $I_j$ be the interval $(A+2^{-j}, A+2^{1-j}]$, $j \in \mathbb N$. Then each interval is non-overlapping, and $$\bigcup_{j=1}^\infty I_j = (A, A+1].$$
Let $I_{j_k}$ be the ordered, non-empty intervals (intervals containing terms of $(a_n)$). Let $a_{n_k}$ be any one term contained in $I_{j_k}$. Then $a_{n_k}\gt a_{n_{k+1}}$, so that $(a_{n_k})$ is a monotone decreasing subsequence of $(a_n)$.
I'm sure there's a much simpler, more elegant way to do this. But does my way work? Is there anything I'm overlooking?

A: Suppose there are infinitely many terms in $(A,\infty).$ Choose $n_1$ so that $a_{n_1}\gt A.$ Then choose $n_2\gt n_1$ so that $a_{n_1}\gt a_{n_2}\gt A.$ Then choose $n_3\gt n_2$ so that $a_{n_2}\gt a_{n_3}\gt A.$ And so on.
