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.
It is not hard to prove that if a Cauchy sequence has a convergent subsequence, then the whole sequence converges. So, suppose that $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence that doesn't converge. If $x\in K$, then $x$ is not the limit of a a subsequence of $(x_n)_{n\in\mathbb N}$. So, there is an open set $A_x$ such that $x\in A_x$ and such that $A_x$ contains only finitely many terms of the sequence. Note that $K=\bigcup_{x\in K}A_x$; so, $(A_x)_{x\in K}$ is an open cover of $K$. Therefore, it hs a finite subcover $(A_{x_j})_{j\in\{1,2,\ldots,N\}}$. But since each $A_{x_j}$ contains finitely many $x_n$'s, this is impossible. A contradiction has been reached and so the sequence $(x_n)_{n\in\mathbb N}$ converges.