How would I go about proving such a statement? Any help would be great thanks.
First prove this theorem: Every sequence contains a monotone subsequence.
Then prove that Every bounded non-increasing (non-decreasing) sequence converges to its least upper bound (greatest lower bound).
Suppose that $\displaystyle \left( {x_n } \right)$ is a sequence.
Define a set $\displaystyle S = \left\{ {n:\left( {\forall j > n} \right)\left[ {x_j > x_n } \right]} \right\}$.
Take notice of this critical fact: $\displaystyle t \notin S\text{ if and only if }\left( {\exists K > t} \right)\left[ {x_K \leqslant x_t } \right]$.
There are two cases to consider: $\displaystyle S$ is infinite.
In this case there is an increasing subsequence.
If $\displaystyle S$ is finite then there is a non-increasing subsequence.
Well, that would be trivial! Just add some numbers to your monotone sequence! For example, the monotone sequence $\displaystyle \{x_1, x_2, x_3, ...\}$ is a subsequence of sequence $\displaystyle \{x_1, 1, x_2, 1, x_3, 1, ...\}$
But what you want is the other way- that every sequence contains a monontone subsequence. I'll give you a start: Let S be the set of all indices i such that if j> i, then $\displaystyle a_i< a_j$. In other words, i is in S if and only if $\displaystyle a_i$ is less than all succeeding numbers in the sequence. Now, consider two cases:
1) S is infinite.
2) S is either finite or empty.