Studying Rudin's Real Analysis and after proving the theorem "Closed subsets of compact sets are compact" I have came up with the following thought: The theorem means that if a set is closed and has a compact superset, then it is compact. This also means that:
A set $K$ is not compact $\implies$ $K$ is not closed or $K$ has no compact superset.
Assume that $K$ is not compact and $K$ is closed. Then it must be that $K$ has no compact superset. I wanted to show that. Trivially, if for every $E$, which is $ E \subseteq K$, $E$ is not compact, then since $K \subseteq K$, $K$ is automatically not compact. While this looks valid, it seemed too trivial to be true to me. Is this really valid? If not, how can one show that when $K$ is not compact, but closed, it has no compact supersets?