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 $N\geq N_{\varepsilon}$, $0\leq \operatorname{diam} E_N< \varepsilon.$
Well you can start by redefining the concept "compact" by stating that a space $X$ is compact if for every family $(F_{\alpha})_{\alpha\in A}$ of closed sets that has the so-called "finite intersection property" the intersection $\bigcap_{\alpha\in A}F_{\alpha}$ is not empty.
Here family $(F_{\alpha})_{\alpha\in A}$ has by definition this "finite intersection property" if for every finite $B\subseteq A$ the intersection $\bigcap_{\alpha\in B}F_{\alpha}$ is not empty.
Using this underlying definition of "compact" (equivalent with the definition that states that open covers must have finite subcovers) it is enough to prove here that in metric spaces every compact subspace $K$ is closed (so that the family of compact sets $(K_{\alpha})$ can be recognized as a family of closed sets).
edit:
The following statements are equivalent:
- (1) Every collection of open sets that cover $X$ has a finite subcover.
- (2) Every collection of closed sets that has the finite intersection propery has a non-empty intersection.
(1)$\implies$(2)
Let $(F_{\alpha})_{\alpha\in A}$ be a collection of closed sets that has the finite intersection property.
Now define $U_{\alpha}=F_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that for every finite $B\subseteq A$ we have $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}\neq X$$
This tells us that the collection $(U_{\alpha})_{\alpha\in A}$ has no finite subcover of $X$. It is a collection of open sets, so this allows us to conclude that $(U_{\alpha})_{\alpha\in A}$ does not cover $X$.
That means that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}\neq\varnothing$$as was to be shown.
(2)$\implies$(1)
Let $(U_{\alpha})_{\alpha\in A}$ be a collection of open sets that cover $X$.
Now define $F_{\alpha}=U_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}=\varnothing$$
That implies the existence of a finite $B\subseteq A$ such that $\bigcap_{\alpha\in B}F_{\alpha}=\varnothing$ (non-existence of such finite $B$ should contradict that closed family $(F_{\alpha})_{\alpha\in A}$ has the finite intersection property).
Then $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}=\varnothing^{\complement}=X$$
Proved is now that cover $(U_{\alpha})_{\alpha\in A}$ has a finite subcover $(U_{\alpha})_{\alpha\in B}$ and we are ready.
Best Answer
I don't have "Baby Rudin" handy at the moment, but if $\bigcap_{n=1}^\infty K_n$ were to contains two distinct points, $x, y$, then $\mathrm{diam} ( \bigcap_{n=1}^\infty K_n ) \geq d (x,y) > 0$. As $\bigcap_{n=1}^\infty K_n \subseteq K_n$ for all $n$, it would follow that $\mathrm{diam} (K_n) \geq d(x,y)$ for all $n$, contradicting that $\mathrm{diam} (K_n) \rightarrow 0$.