Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,q\in E$. The sup of $S$ is called the diameter of $E$, and written $\text{diam }E$.
Theorem. In $\mathbb{R}^k$, every Cauchy sequence converges.
Let $E_N$ be the sequence of the points $\boldsymbol{x}_N, \boldsymbol{x}_{N+1}, \dots$
In some part of the proof, they use that for some $N$, $\text{diam } E_N < 1$ and that the range of a Cauchy sequence $\{\boldsymbol{x}_n\}$ is the union of $E_N$ and the finite set $\{\boldsymbol{x}_1,\dots,\boldsymbol{x}_{N-1}\}$, all of which I can see it's true, to conclude that $\{\boldsymbol{x}_n\}$ is bounded, which I can't understand.
Please let me know if I have misunderstood your issue with the proof. From what you are saying it sounds like the issue is how Rudin concludes that the Cauchy sequence $\{x_n\}_{n=1}^\infty$ is bounded from \begin{align} &\text{(i) } \text{ there is a positive integer } N \text{ so that } E_N \text{ is bounded } (\text{diam } E_N < 1) \\& \\& \text{ and} \\& \\&\text{(ii) } \text{ the complement of } E_N \text{ in } R\left[\{x_n\}\right]:=\{ p \in \mathbb{R}^k : p \in E_1\} \text{ is a finite set of numbers in } \mathbb{R}^k \\&\left(R\left[\{x_n\}\right]\setminus E_n = \{x_1, x_2, \ldots, x_{N-1}\}\right). \end{align} In $\mathbb{R}^k$, the finite union of bounded subsets is bounded. So (i) and (ii) combine to show that $\{x_n\}$ is bounded because its range \begin{equation} R\left[\{x_n\}\right]=E_N \cup R\left[\{x_n\}\right]\setminus E_n \end{equation} is a bounded set.
edit (just read comments under your question):
From (i) we know that $E_N$ is bounded because $r \in E_N$ implies \begin{align} d\left(0,r\right) & \leq d\left(0, x_N\right) + d\left(x_N, r\right) \\& < d\left(0, x_N\right) + 1\, . \end{align}