I'd like to prove following proposition (all definitions below are taken from Terence Tao's Analysis I ):
Proposition. If $(a_{i})_{i=1}^{\infty}$ is a Cauchy sequence that is bounded away from zero, then there must exist some $N≥1$ such that the subsequence $(a_{i})_{i=N}^{\infty}$ is either positively or negatively bounded away from zero.
My attempt:
We will use contradiction. Take some Cauchy sequence $(a_{i})_{i=1}^{\infty}$ that is bounded away from zero (i.e., there exists some $c > 0$ such that $|a_i| > c$ for all $i ≥ 1$), and suppose that there is no subsequence $(a_{i})_{i=N}^{\infty}$ that is either positively or negatively bounded away from zero. It follows that for arbitrary $N ≥ 1$, there must be at least one $i ≥ N$ such that $a_i > c$ and at least one $j$ such that $a_j < -c$. Now we will show that $(a_{i})_{i=1}^{\infty}$ cannot be a Cauchy sequence, obtaining a contradiction.
Take some $0 <\epsilon < c$ and take arbitrary $N ≥ 1$. As stated above, we can find some $i ≥ N$ such that $a_i > c$ and some $j$ such that $a_j < -c$. Then we have $-a_j > c$, and thus
$$a_i - a_j > 2c$$
Note that $a_i - a_j > 0$, which means
$$|a_i - a_j| > 2c > \epsilon$$
Since $N$ was arbitrary, it follows that $(a_{i})_{i=1}^{\infty}$ is not eventually $\epsilon$-steady, and thus $(a_{i})_{i=1}^{\infty}$ cannot be a Cauchy sequence, which contradicts our initial assumption.
$\Box$
Is it correct?