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.

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.
