Cauchy sequences and diameter of a set The definitions and theorem below are from baby Rudin before theorem 3.10.

Definition. A sequence $\{p_n\}$ in a metric space $X$ is said to be a Cauchy if for every $\epsilon > 0$ there is an integer $N$ such that $d(p_n, p_m) < \epsilon$ if $n \ge N$ and $m \ge N$.
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 $E$.
Theorem. Let $\{p_n\}$ be a sequence and $E_N$ consisting of the points $p_N, p_{N+1}, p_{N+2}, \dots$, then $\{p_n\}$ is a Cauchy sequence $\iff$ $\lim\limits_{N\to \infty} \mbox{diam}\quad E_N = 0$.

Ok. If a sequence is Cauchy, then for any $n,m \ge N$, $$d(p_n, p_m) < \epsilon, $$ so any element of $S=\{ d(p_z, p_w), z,w \ge N\}$ will be lesser than this $\epsilon$. But $\epsilon$ is arbitrary, so all elements of $S$ will be zero.
So the sup of $S$, that is, the diameter of $E_N$ will be zero, that's right?
To do the $(\Leftarrow)$ part, what does $\lim\limits_{N\to \infty} diam E_N = 0$ exactly means? Using the strict definition given in the book, we have that, given a $\epsilon > 0$, there is a $N$ such that the distance between a term with index greater than $N$ and the 0 (in this case) is lesser than $\epsilon$.
But I don't know how to interpret it because the term here is the sup of a set $E_N$, and I don't know how it will be arranged in a sequence, so that the definition on limit makes sense.
 A: For $\boxed{\Leftarrow}$:
For any $N$, you have $E_N = \{p_N, p_{N+1}, p_{N+2}, \dots,\}$. We have the assumption that $\lim_{N\to\infty}\operatorname{diam} E_N=0$ (to understand what it means, note that $(\operatorname{diam} E_N)_{N\geq 0}$ is a sequence of real numbers); that is,

For every $\varepsilon > 0$ there exists $N_\varepsilon \geq 0$ such that, for every $N\geq N_\varepsilon,$ $\operatorname{diam} E_N \leq \varepsilon$.

So fix any $\varepsilon > 0$, and consider the corresponding $N_\varepsilon$. Fix any $n\geq N_\varepsilon$ and $m\geq N_\varepsilon$; we have that $p_n,p_m\in E_{N_\varepsilon}$ (by definition of $E_{N_\varepsilon}$).
Therefore, $$d(p_n,p_m)\leq \sup_{x,y\in E_{N_\varepsilon}} d(x,y) = \operatorname{diam} E_{N_\varepsilon} \leq \varepsilon$$
showing that the sequence is Cauchy.

Now, let us turn to the other direction, $\boxed{\Rightarrow}$. You wrote:

Ok. If a sequence is Cauchy, then for any $n,m \ge N$, $$d(p_n, p_m) < \epsilon, $$ so any element of $S=\{ d(p_z, p_w), z,w \ge N\}$ will be lesser than this $\epsilon$. But $\epsilon$ is arbitrary, so all elements of $S$ will be zero.


So the sup of $S$, that is, the diameter of $E_N$ will be zero, that's right?

This is the overall idea, but is a bit sloppy, and the conclusion is not quite correct (all elements of your set will not be zero -- zero is not actually necessarily an element of your metric space! But all elements of $S$ will have to be very close to each other: this is what the diameter of a set captures.
To show the sequence of diameters of the sets converges to zero, let us do it the "hard way." Fix any $\varepsilon > 0$. By our assumption that the sequence $(p_n)_n$ is Cauchy, there exists some $N_\varepsilon > 0$ such that $d(p_n,p_m) < \frac{\varepsilon}{2}$ forall $n,m\geq N_\varepsilon$.
Consider now any $N\geq N_\varepsilon$: we want to show that that $\operatorname{diam} E_N < \varepsilon$. By definition of the siameter as the supremum, there exist $x,y\in E_N$ such that
$$
0\leq \operatorname{diam} E_N = \sup_{x',y'\in E_N} d(x',y') < d(x,y) + \frac{\varepsilon}{2}.$
$$
But by definition of $E_N$, $x$ and $y$ are of the form $P_n$ and $p_m$ respectively, for some $n,m\geq N\geq N_{\varepsilon}$. By our choice of $N_{\varepsilon}$, this implies that $d(x,y) < \frac{\varepsilon}{2}$, and therefore
$$
0\leq \operatorname{diam} E_N < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
$$
This is enought to conclude, as $\varepsilon>0$ was arbitrary:

For every $\varepsilon>0$, there exists $N_{\varepsilon}$ such that, for every $N\geq N_{\varepsilon}$, $0\leq \operatorname{diam} E_N< \varepsilon.$

