Question: How to prove that the sequence $\{a\_n\}$ has no convergent subsequence if and only if $|a\_n|$ approaches infinity?
The forward direction is obvious, how to prove the other direction?

Answer: 
Use these two facts:

* Every sequence has a monotone subsequence.
* Every bounded monotone sequence converges.
As others have stated, you need a theorem of Bolzano (proved below), usually known as the Bolzano$-$Weierstrass theorem, that a bounded sequence has a *cluster point*: a point such that an infinite number of terms of the sequence lie within any given neighbourhood of the point. First note that, if the terms do not approach $\infty$ in magnitude, then infinitely many of them must be bounded away from $\pm\infty$ and so lie within a finite interval, say $[a\,,b]$, with $a<b$.

For convenience, scale the terms by $x\mapsto (x-a)/(b-a),$ so that we are dealing with infinitely many terms in $[0\,,1].$ Cut this interval in half. Then at least one half has infinitely many terms. Choose such a half, and record $s\_1=0$ if it is the left half or $s\_1=1$ if the right half. Cut that reduced interval in half and, as before, choose a half with infinitely many terms, and record a corresponding $s\_2=0$ or $s\_2=1$ again. Continue this process indefinitely, to produce an infinite nest of intervals, each being half of its predecessor and containing infinitely many terms of the original scaled sequence, along with a corresponding string $s\_1s\_2\dots$ of $0$s and $1$s. Let $s$ be the real number represented as $\cdot s\_1s\_2\dots$ in binary notation. Then, by reverting to our original number scale, the required cluster point is $a+(b-a)s$.