I have proved the following statement and would like to know if my proof seems correct.
Let ε > 0. Show that if $(a_n)$ and $(b_n)$ are eventually ε-close, then $(a_n)$ is bounded iff $(b_n)$ is bounded.
see this snapshot from the text itself
Proof: Since $a_n$ and $b_n$ are ε-close, we know that for all $ε > 0$ there exists a natural number $N$, such that $$|a_n - b_n| \le ε,$$ $$ n\ge N $$ Since $b_n$ is bounded, $|b_n|<=M_1$ for some natural number $M_1$. Set $ε = 1$. Then there exist a natural number $N$ such that $$|a_n - b_n| \le 1,$$ $$ n\ge N $$ Applying the reverse triangle inequality: $$ |a_n| - |b_n| \le |a_n - b_n| \le 1$$ thus: $$ |a_n| \le 1 + |b_n| $$ For $ n < N $, $ (a_n)$ is a sequence of finite elements and thus bounded (it is assumed this fact does not need to be proved), say by the natural number $M_2$. For $ n \ge N, |a_n| \le 1 + M_1$ (since $b_n$ is bounded).
Set $M_3 = max(M_2, 1 + M_1)$. Then for all $n$, $|a_n| \le M_3$.
Thus $(a_n)$ is bounded, and the proof is completed.