Prove that if $a_n$ contains no convergent subsequences, then |$a_n$| approaches infinity, as n approaches infinity.

I first approached the problem using contradiction.

Suppose that |$a_n$| does not approach infinity. Case 1: The sequence will converge to some L, and thus be bounded. Case 2: The sequence will neither converge nor diverge, thus it will be bounded. (I'm having trouble explaining case 2)

For both cases, we can use the Bolzano-Weiestrass Theorem, thus |$a_n$| has a convergent subsequence.(How can I show that $a_n$ will also have a convergent subsequence?)

This is a contradiction to the given statement, thus |$a_n$| will approach infinity.

This was how I approached the problem, but there's a few problems with it. Please let me know if there is a better way or how I can improve mine.

Thank you in advance!

OR

Again using contradiction, I can say that there exists an M > 0 for all N in the natural numbers, where there exists n > N such that we have |$a_n$| ≤ M.

Let n = $a_ni$. Then |$a_ni$| ≤ M. Thus -M < $a_ni$ < M. This sequence is bounded, thus, by the Bolzano-Weiestrass theorem, it has convergent subsequences. CONTRADICTION