In "Principles of Mathematical Analysis" by Walter Rudin, Theorem 2.33 asserts that compactness is a universal property: meaning if $K \subset X$ is compact in $X$, then $K$ is compact in any embedding metric space. Theorem 2.34 asserts that "compact subsets of metric spaces are closed": meaning for any $K \subset X$, if $K$ is compact, then $K$ is closed relative to $X$.
Does this mean (by taking the contrapositive) that for any set $E$, if we can find just one metric space $X$ such that $E \subset X$ and $E$ is not closed relative to $X$, then $E$ is not compact anywhere? (This would make it easy to prove, for example, that no segment $(a, b)$ is compact...)
OR should the contrapositive be read: if $E$ is not closed relative to every embedding metric space $X$, then $E$ is not compact anywhere? (This would make the contrapositive relatively useless, since every set is closed relative to itself...)
Definitions: Let $X$ be a metric space with distance function $d(p, q)$. For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) < r\}$. Any $p \in X$ is a limit point of $E$ if $\forall r > 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$. Any subset $E$ of $X$ is closed if it contains all of its limit points. For any subset $E$ of $X$, an open cover is a collection of sets $\{G_\alpha\}$ which are open in $X$, such that $E \subset \bigcup_\alpha G_\alpha$. $E$ is compact if every one of its open covers has a finite open subcover.