Show that if $x_n$ has no convergent subsequences then $|x_n| \to \infty$ Suppose that the sequence $x_n$ has no convergent subsequences. Let M $ > 0$. 
Prove that there exists at most finitely many values of $n$ such that $x_n \in [-M,M].$ Explain why this implies |$x_n$|$\rightarrow$$\infty$ as $n \to \infty$
This is part a of 1.62 from Advanced Calculus by Leonard Richardson. 
I have no idea where to start with this proof. 
 A: We argue by contradiction. Assume that there are infinitely many values of $n$ such that $x_n \in [-M,M]$ then take a subsequence $x_{n_k}$ such that $x_{n_k} \in [-M,M]$ for all $k$.  Then $x_{n_k}$ is bounded and must have a convergent subsequence by the Bolzano-Weierstrass theorem. So $x_n$ must have a convergent subsequence. Contradiction. 
The second part follows from using the definition of what it means for a sequence to go to $\infty$. 
A: So, you can use some standard results from analysis to deduce that, but one could argue they obscure just how simple this question is once you have the right point of view.
If there are infinitely many terms of the sequence in $[-M,M]$, then it's quite obvious there are infinitely many in either $[-M,0]$ or $[0,M]$. Say it's $[0,M]$ WLOG; then continuing there are infinitely many in either $[0,M/2]$ or $[M/2,M]$, and so on. This defines an infinite nested sequence of intervals that will converge to some number (the intersection of all the intervals), and this number is the limit of a convergent subsequence of the $x_i$ (take one term from each interval). So this shows the first part, and the second part now follows easily :)
