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The original question proof is "Every bounded monotone sequence has a limit."

I've seen a lot of proofs of this that make use of the least upper bound, but that's not something we're permitted to use. Instead, we're to show this through the contrapositive of the statement "every bounded monotone sequence is a Cauchy sequence."

So ultimately I want to show that a monotone sequence that is not Cauchy cannot be bounded.

If it's monotone, then $s_{n+1} \leq s_n$ for all $n$

If it's unbounded, then $s_n \geq A$ for all $n$ greater than some $N$ and for all $A$

And if it's not Cauchy, then $\exists \epsilon > 0 \ \forall N \ \exists m,n \ |s_m - s_n| \geq \epsilon$

I'm just having trouble putting these together to form an argument. Any help is appreciated.

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    How are you not allowed to use completeness? This statement is not true in a non complete space. – 2017-02-28
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    The completeness we're to use is that "Every Cauchy sequence of real numbers converges to a real number". We're only to avoid using the supremum since we haven't proved that it exists for every bounded monotone sequence yet – 2017-02-28

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To prove it this way notice the order of the quantifiers in the last statement. $\exists \epsilon > 0 \ \forall N \ \exists m,n \ |s_m - s_n| \geq \epsilon$

There is one epsilon that does not depend on anything else. Let $\epsilon$ be this epsilon. Pick $N_1 = 1$ then we know there are $m,n$ such that $|s_m - s_n| \geq \epsilon$. The let $N_2 = max(m, n) + 1$ so we have to pick out two more $m,n$.

Now we want to show that $s_n$ is unbounded. So given some $M$ we can repeat the above process $\lceil M/\epsilon \rceil$ times which means that $s_n$ will be larger than $M$. So $s_n$ is unbounded.

Note that this proof needs a few details filled in and some better wording but you can do that on your own. Specifically you should show that this guarantees that this is actually growing. One nice way to do this would be to pick out a subsequence by the above process.