Baby Rudin 3.11 (b) I hope I could get clarification on a minor detail in the proof to theorem 3.11 (b) in Rudin's Principles of Mathematical Analysis. The theorem and proof are as follows.
Theorem:

If X is a compact metric space and if {$p_n$} is a Cauchy sequence in X, then {$p_n$} converges to some point of X.

Proof:

Let {$p_n$} be a Cauchy sequence in the compact space X. For N = 1,2,3..., let $E_N$ be the set consisting of $p_N$, $p_{N+1}$, $p_{N+2}$, ... Then $$ \lim \limits_{N \to \infty} diam \overline {E_N} = 0$$, by Definition 3.9 and Theorem 3.10 (a). Being a closed subset of the compact space X , each $\overline {E_N}$ is compact (Theorem 2.35). Also $E_N \supset E_{N+1}$, so that $\overline{E_N} \supset \overline{E_{N+1}}$...

Here I am trying to find out a reasoning behind the statement $E_N \supset E_{N+1}$, then $\overline{E_N} \supset \overline{E_{N+1}}$. My reasoning behind $E_N \supset E_{N+1}$ is that since $ \lim \limits_{N \to \infty} diam E_N = 0$, and since the diameter of $E_N$ captures how big the set is, then as N gets bigger, the set becomes smaller. And the reason why $\overline{E_N} \supset \overline{E_{N+1}}$ is that $diam \overline{E_N} = diam E_N$ (theorem 3.10 a). Is my reasonings correct?
 A: This is actually a general property of metric spaces (and topological spaces, more generality): if $A \subset B$ then $\bar{A} \subset \bar{B}$, where $\subset$ means subset, not necessarily proper. The proof for metric spaces goes as follows: If $x \in \bar{A}$, there is a sequence $a_n \to x$ for $a_n \in A$. Since $A \subset B$, the sequence $a_n$ is also in $B$, so $x$ is also a limit point of $B$, so $x \in \bar{B}$.
A: A simpler proof, I think! Fix an $\epsilon>0$ then there exists an $M$ such that for $n,m\geq\,M$ we have $|x_{n}-x_{m}|<\dfrac{\epsilon}{2}$. But $X$ is compact and hence there is a subsequence of $x_{n}$ say $x_{n_{k}}\to \bar{x}$ in $X$. So for $k$ sufficiently large we get $|x_{n_{k}}-x_{m}|<\dfrac{\epsilon}{2}$ and by continuity of the absolute value we get $|\bar{x}-x_{m}|\leq\dfrac{\epsilon}{2}<\epsilon$.
Therefore for a fixed $\epsilon$ we have found an $M$ such that $m\geq\,M$ implies that $|x_{m}-\bar{x}|<\epsilon$ which is the definition of $x_{m}\to\bar{x}$
