I'm self studying Principles of Mathematical Analysis by Walter Rudin (so I welcome any feedback), and I'm wondering if there is any fault to this self written proof, which I think is simpler than the one presented by Rudin.
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:
By theorem 3.6(a), there exists a sequence $\{n_j\}$ of positive integers $n_1<n_2<n_3<\cdots$ ($j=1,2,3,\ldots$) such that $\{p_{n_j}\}$ converges to some point $p$ in $X$. Fix $\varepsilon>0$. Since $\{p_n\}$ is a Cauchy sequence, there exists some integer $N$ such that $n\geq N$ implies $d(p_n,p_N)<\frac{\varepsilon}{3}$. Also, for all but finitely many $j$, $d(p_{n_j},p)<\frac{\varepsilon}{3}$, so choose any such $j$ which satisfies the condition $j\geq N$. Then $n_j\geq N$ (since $n_j\geq j$) so $d(p_{n_j},p_N)<\frac{\varepsilon}{3}$. Finally, $$d(p_n,p)\leq d(p_n,p_N) + d(p_N,p_{n_j}) + d(p_{n_j},p)<\varepsilon.$$ Hence, $p_n \rightarrow p$.
Best Answer
Your proof is totally correct. The only reasonable "improvement" I see is explicitly stating that the last inequality holds whenever $n \ge \text{max}\{N, N_0\}$, where $N_0$ is such that $n_j \ge N_0$ implies $d(p_{n_j}, p) < \frac{\epsilon}{3}$. Then it is perfectly clear right away after which element the inequality is true.