I've seen other proofs of this on here but all of them seem to rely on showing compact iff sequentially compact in a metric space but instead I want to use only the definition of compactness, that every open cover of a compact space admits a finite subcover to prove this.
Proof by contradiction.
Suppose K is compact and not complete. Then there is a Cauchy sequence in K that is not convergent in K. Let $(x_n)$ be this sequence. I want to somehow show that this will give an open cover with no finite subcover but I can't see how I do that. Or I want to show that somehow the cluster point for the cauchy sequence must be in K. But I can't see how I can show that.