$\color{blue}{\text{Required Definitions and Results:}}$
Proposition 4.3.3 (b). Let $x, y$ be rational. We have $|x + y| \leq |x| + |y|$.
Proposition 4.3.3 (f). $|x-y| = |y-x|$.
Definition 5.1.12 (Bounded sequences). Let $M \geq 0$ be rational. A finite sequence $a_1, a_2, \ldots, a_n$ is bounded by $M$ iff $|a_i| \leq M$ for all $1 \leq i \leq n$. An infinite sequence $(a_n)_{n=1}^{\infty}$ is bounded by $M$ iff $|a_i| \leq M$ for all $i \geq 1$. A sequence is said to be bounded iff it is bounded by $M$ for some rational $M \geq 0$.
Lemma 5.1.14 (Finite sequences are bounded). Every finite sequence $a_1, a_2, \ldots, a_n$ is bounded.
Definition 4.3.4 ($\varepsilon$-closeness). Let $\varepsilon > 0$ be a rational number, and let $x, y$ be rational numbers. We say that $y$ is $\varepsilon$-close to $x$ iff we have $|y-x| \leq \varepsilon$.
Definition 5.2.3 (Eventually $\varepsilon$-close sequences). Let $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ be two sequences, and let $\varepsilon > 0$. We say that the sequence $(a_n)_{n=0}^{\infty}$ is eventually $\varepsilon$-close to $(b_n)_{n=0}^{\infty}$ iff there exists an $N \geq 0$ such that the sequences $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ are $\varepsilon$-close.
Exercise 5.2.2. Let $\varepsilon > 0$. Show that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are eventually $\varepsilon$-close, then $(a_n)_{n=1}^{\infty}$ is bounded iff $(b_n)_{n=1}^{\infty}$ is bounded.
My proof.
Let $\varepsilon > 0$. Suppose that $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are eventually $\varepsilon$-close. This implies there exists a natural number $N \geq 0$ such that
$$ |a_n - b_n| \leq \varepsilon, \ \text{for all} \ n \geq N \tag{1} $$
Suppose $(a_n)_{n=1}^{\infty}$ is bounded. This implies there is some rational $M_1 \geq 0$ such that
$$ |a_j| \leq M_1, \ \text{for all} \ j \geq 1 \tag{2} $$
We have that
$$ |b_j| = |b_j - a_j + a_j| \leq |b_j - a_j| + |a_j| \leq |a_j - b_j| + M_1 \tag{3} $$
where the first inequality comes from Proposition 4.3.3(b) and the second inequality comes from Proposition 4.3.3(f) and $(2)$.
For $j = n \geq N$, we have that
$$ |b_n| \leq |a_n - b_n| + M_1 \leq \varepsilon + M_1 $$
where the first inequality comes from $(3)$ and the second inequality comes from $(1)$. This shows that the infinite sequence $(b_n)_{n=N}^{\infty}$ is bounded by $\varepsilon + M_1$.
This leaves us with the finite sequence $b_1, b_2, \ldots, b_{N-1}$. By Lemma 5.1.14, this sequence is bounded, implying there exists a rational $M_2 \geq 0$ such that $|b_j| \leq M_2$ for all $j \leq N-1$.
Thus, by the discussion above, we must have
$$ |b_j| \leq M_1 + M_2 + \varepsilon, \ \text{for all} \ j \geq 1 $$
Denoting $M_3 := M_1 + M_2 + \varepsilon$, we have shown $(b_n)_{n=1}^{\infty}$ is bounded. The converse is shown by interchanging the roles of $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$.
My Question.
Assuming my proof is correct, I feel that the bound that I do end up with is larger than needed.
For example, if we have a sequence $(a_n)_{n=1}^{\infty}$ defined by
$$ 0, 1, 0, 1, 0, \ldots $$
and $(b_n)_{n=1}^{\infty}$ defined by
$$ 2, 0, 2, 0, 2, \ldots $$
We see that the two sequences are eventually $2$-close, i.e.,
$$ |a_n - b_n| \leq 2 $$
and cannot be any eventually closer. Also, we see that $(a_n)_{n=1}^{\infty}$ is bounded by $1$, i.e.,
$$ |a_j| \leq 1 $$
for all $j \geq 1$. Thus, by my proof, we must have that $(b_n)_{n=1}^{\infty}$ is bounded by $M_1 + 1 +2 = M_1 + 3$ where $M_1 \geq 0$ is some rational, i.e.,
$$ |b_j| \leq M_1 + 3 $$
Now, while this is true, we can clearly see that we can do better, that is we have that $|b_j| \leq 2$ for all $j \geq 1$ is true, but my proof hasn't constructed this case. Why is that? Is there another proof that can construct such a bound?