Prove that every unbounded sequence contains a monotone subsequence that diverges to inifnity. I think I have the basic framework for this proof, but I am having trouble putting everything together in a convincing way.  
I plan on showing that by the definition, an unbounded sequence $(a_n)$ has infinitely many $n$ such that $a_n>M$ where $M>0$, for some natural number $n$. I'm not sure how it is implied that there exists some $a_{n_k}$ such that $a_{n_k}>M$ for some natural number $k$.  Let me know if I'm on the right track, and please help me string these ideas together!
 A: Assuming that the sequence is unbounded above, then you can generate your monotone divergent sequence as follows.
Choose $n_1$ to be the first index such that $a_{n_1} > 1$.  Now, from the remaining sequence, choose $n_2 > n_1$ to be the first index such that $a_{n_2} > 2$ and $a_{n_2} > a_{n_1}$.
Rinse.  Repeat.
For each $k \in \mathbb{N}$, you have $a_{n_k} > k$ and $a_{n_k} > a_{n_{k-1}}$, so the subsequence $\left( a_{n_k} \right)$ certainly diverges to infinity.
Why do you know that you can always find such an index?  If you could not, then your sequence must have been bounded.
A: You surely should use that there are infinitely many $n$'s such that $|a_n| > M$ for any real $M$. Consider one of positive and non-positive subsequences. At least one of them is not bountded, too. So you have right to assume without lose of generality that there are infinitely many $n$'s such that $a_n > M$ for any real $M$. If follows that there are infinitely many such $n$'s greater any $n'$. And so you may choose any $a_i$ and let $M = a_i$ to find $j$ such that $j>i$ and $a_j > a_i$. Do you see how to get monotonic subsequence converging to $+\infty$?
A: I believe it's because you'd like to guarantee the convergence of $a_{n_k}$ to $\infty$. Actually, you could use the terms of any divergent sequence $b_k$ to do that comparation. 
For example, you could chose $n_k$ to guarantee that $a_{n_k}>k^2$. The reasoning would be the same, because $b_k = k^2 \rightarrow \infty$ when $k \rightarrow \infty $
