Basically, Rudin states that a Cauchy sequence in a compact metric space is convergent. However, his proof is a bit non-intuitive and different from what one comprehends at first sight.
Thus, I have put my own proof, but I want it to be checked if correct, since it seems to be way more intuitive.
If $p_n$ takes only a finite number of values but is not ultimately constant, then it is obvious that cannot be Cauchy. Thus, a Cauchy sequence has an infinite number of values.
Thus, a Cauchy sequence has a convergent subsequence $p_{n_k}$ that converges to $p$. Thus, there is $N_1$ such that $d(p_{N_1},p)<\frac{\varepsilon}{2}$
However, since it's a Cauchy sequence, $d(p_n,p_m)<\frac{\varepsilon}{2}\forall n,m>N_2$ for some $N_2<N_1$. (To obtain that, we choose a large enough $N_1$)
Thus, $\forall n>N_2$, $d(p_n,p)\leq d(p_n,p_{N_1})+d(p_{N_1},p)<\varepsilon$.
QED.
The statement "a Cauchy sequence has a convergent subsequence $p_{n_k}$ that converges to $p$" uses the result that every sequence in a compact metric space has a convergent subsequence. If you proved this earlier, the proof looks fine.