Cauchy sequence in compact metric space converges; incorrect proof? I'm self-studying real analysis, and I've been trying to prove the following statement:
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 $X$.
My proof is as follows:
Fix $\epsilon>0$. Since $\{p_n\}$ is Cauchy, we know that there exists an integer $N$ such that $m,n\geq N$ implies that $d(p_m,p_n)<\epsilon$. Then $\{p_n\}$ converges to $p_N$ because $n\geq N$ implies that $d(p_n,p_N)<\epsilon$. Since $p_N \in \{p_n\}$, we know that $p_n \in X$.
I am pretty sure my proof is incorrect because it doesn't use the compactness of the metric space anywhere in the argument, but I'm having trouble figuring out where I went wrong in my argument. I have read other proofs of this statement and understand why they work; the issue is figuring out what I've overlooked in my proof.
 A: Your proof is not correct since your $N$ depends on $\epsilon.$ Your $p_N$ may not work for other $\epsilon,$ since you'll find another $p_N$ if you change $\epsilon.$
Let $\{p_n\}$ be a Cauchy sequence in $X$. Since $X$ is a compact metric space, it is sequentially compact, so there exists a convergent subsequence $\{p_{n_k}\}$ of $\{p_n\}$ such that $p_{n_k}$ converges to some $p \in X.$ 
We show that $p_n \to p$ and we are done.
Let $\epsilon>0.$ Since $p_{n_k}\to p,$ there exists $N_1 \in \mathbb{Z}_+$ such that $|p_{n_k}-p|<\frac{\epsilon}{2}$ for all $n_k \geq N_1.$ Since $\{p_n\}$ is cauchy, there exists $N_2 \in \mathbb{Z}_+$ such that $|p_n-p_m|<\frac{\epsilon}{2}$ for all $n,m \geq N_2.$ Now let $N=\max\{N_1,N_2\}.$ Thus, if $n>N$ then
$$ |p_n-p|\le |p_n-p_N|+|p_N-p|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$
so $p_n \to p.$
A: You did not prove that that it converges to $p_N$. Convergence to $p_N$ means that for every $\varepsilon>0$, you have $d(p_N,p_n)<\varepsilon$ for all $n$ large enough. But when you proved the inequality $d(p_N,p_n)<\varepsilon$, you proved it for a fixed $\varepsilon$, not for all of them.
