Proof of Rudin's Theorem 3.11 (c) 
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.
 A: If you 'add' a finite set to a bounded set in Euclidean space, then that set is also bounded. 
Also: Every finite set is bounded and the union of two bounded sets is bounded.
Addressing student's direct problem mentioned in a comment:
Rudin's Definition (2.18)-(i) for a bounded metric spaces is OK and has nothing to do with an open ball about the $0 \text{ vector}$ in Euclidean space. Right before Theorem 3.11, Rudin provides Definition 3.9 for the diameter of a metric space. It looks like he never mentions the following fact:
Proposition: Let E be any metric space. E is bounded iff it has a finite diameter. 
Proof: If E is bounded there is a point $q$ in $E$ and a number $M$ so that all points in $E$ are in the 'open ball' of 'radius' $M$ about $q$. By using the triangle inequality, you can easily see that the diameter is less than or equal to $2M$.
If E has a finite diameter, say $D$, choose any point point $q$ in $E$. By the definition of diameter, all points are in the 'open ball' of 'radius' $D + \frac{1}{1000000}$ about $q$, so E is bounded.
QED
Some analysts define bounded to mean finite diameter.
A: 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}
