Do non-compact spaces always contain "big" closed subsets?

Yes - this largely follows from the the definition of "compact". In particular, suppose that $X$ is not compact. This is equivalent to saying that there exists some open cover $\mathscr U$ with no finite subcover. Let $U\in\mathscr U$ be any non-empty set in this cover. The complement $X\setminus U$ cannot be compact, because then there would be a finite subset $\mathscr U'\subseteq \mathscr U$ that covered $X\setminus U$ and then $\{U\}\cup \mathscr U'$ would be a finite subcover of $\mathscr U$, contradicting the hypothesis.

Let $\{U_\alpha\}$ be an open cover of $X$ by nonempty open sets with no finite subcover, and pick an arbitrary $U$ from this cover. Then $C=X-U$ is a noncompact proper closed subset of $X$. The same cover of $C$ has no finite subcover for otherwise we could add $U$ to form a finite subcover of $X$.