I am required to prove the following Lemma. What follows is my attempt can someone please verify that it is correct.
Please consider the following Definitions and Lemma prior to reading the Argument.
Definition. A Sequence $(a_n)_{n=0}^{\infty}$ of rational numbers is eventually $\epsilon$ steady if and only if there exists an $N\in\mathbf{N}$ such that $d(a_j,a_k) = |a_j-a_k|\leq \epsilon$ for all $i,j\ge N$.
Definition. A Sequence $(a_n)_{n=0}^{\infty}$ of rational numbers is a Cauchy Sequence if it is eventually $\epsilon$ steady for all $\epsilon>0$.
Lemma. $5.1.14$ Every finite sequence of rational numbers $(a_n)_{n=0}^{m}$ is bounded.
MY ATTEMPT:
Lemma $5.1.15$ (Cauchy sequences are bounded). Every Cauchy sequence $(a_n)_{n=1}^{\infty}$ is bounded.
Proof. Let $S = (a_n)_{n=0}^{\infty}$ be an arbitrary cauchy sequence. From Definition $S$ is eventually $\epsilon$ steady for every $\epsilon>0$, then in particular for $\epsilon = 1$ we have an $N\in\mathbf{N}$ $\text{s.t.}$ $\forall j\ge N\forall k\ge N(|a_j-a_k|\leq1)$. Consequently we may split the $S$ $\text{s.t.}$ $S = (a_n)_{n=0}^{N-1}+(a_n)_{n=N}^{\infty}$.
For $P = (a_n)_{n=0}^{N-1}$ we have a bound $M$ from Lemma $5.1.14$ and for $Q = (a_n)_{n=N}^{\infty}$ we know that $|a_j-a_N|\leq 1$ for an arbitrary $j\ge N$. Equivalently $-1+a_N\leq a_j\leq 1+a_N$. Now we know from proposition $4.3.3$ that $-|a_N|\leq a_N\leq |a_N|$ from which we may deduce that $-|a_N|-1\leq a_N-1$ and $a_N+1\leq |a_N|+1$ consequently $-1-|a_N|\leq a_j\leq 1+|a_N|$ and thus $|a_j|\leq |a_N|+1$.
Now let $H = M+|a_N|+1$ and let $i\in\mathbf{N}$. If $1\leq i\leq N$ then $|a_i|\leq M\leq M+|a_N|+1 = H$ and similarly for $i\ge N$ we have $|a_i|\leq |a_N|+1\leq M+|a_N|+1 = H$. Therefore for all $i\in\mathbf{N}$ we have $|a_i|\leq H$ thus $S$ is bounded.
$\blacksquare$