Suppose $K$ is compact. What other types of coverings must have finite subcovers? Let $X$ be a topological space. Call $S\subseteq X$ an $\mathcal{O}$-set if there exists an open set $O$ such that $O\subseteq S \subseteq \overline{O}$. Suppose $X$ is compact. Is it true that any covering of $X$ by $\mathcal{O}$-sets has a finite subcovering?
It seems intuitively that it should be true for $X \subseteq \mathbb{R}$, but I'm not sure if this relies on some other topological property of the reals.  
Below the fold I have included some background as to why I would ask such a question, if you're interested.

Background: here on Reddit user posted an interpretation of compactness based on Zeno's paradox--the "steps" that the racer takes are considered as open sets, and a compact "racetrack" implies that the the race must be completed in finite time if the velocity function is continuous. I wondered: why are open sets the correct way to view the steps? Then I got to thinking, a set being non-empty and open is one way to guarantee that it really represents a "chunk" of the space...not just a point, or a subspace of lower dimension (if it were a vector space). But still, why open? I'd rather have the freedom to consider the "step" as any set containing an open set, but contained in its closure. If this is possible, then it becomes clear that this holds for any reasonable interpretation of a "step".
 A: ADDED: There's actually a counterexample way simpler than the one below: $$[0,1]=\left[\frac12,1\right]\cup\bigcup_{n\in\mathbb N}\left[0,\frac12-\frac{1}{4n}\right).$$

ORIGINAL ANSWER: Let $X=$ the Alexandroff one-point compactification of $\mathbb R$, with the extra point denoted as $\omega$. Let $S_0=\{\omega\}\cup(-\infty,0]\cup[1,\infty)$ and $S_n=(0,(n-1)/n)$ for all $n\in\mathbb N$.
Claim$\phantom{---}$ $S_n$ is an $\mathcal O$-set for all $n\in\mathbb N\cup\{0\}$.
Proof:$\phantom{---}$ First note that $S_0$ is closed, because $S_0^c=(0,1)$ is open. Moreover, $U_0=\{\omega\}\cup(-\infty,0)\cup(1,\infty)$ is open because its complement $[0,1]$ is compact in $\mathbb R$. Additionally, neither $\{\omega\}\cup(-\infty,0]\cup(1,\infty)$ nor $\{\omega\}\cup(-\infty,0)\cup[1,\infty)$ is closed (their complements are $(0,1]$ and $[0,1)$, respectively, which are not open), so $\overline{U_0}$ must be actually equal to $S_0$.  It follows that $S_0$ is an $\mathcal O $-set. If $n\in\mathbb N$, then $S_n$ is open in $\mathbb R$ and hence in $X$, so it is obviously an $\mathcal O$-set. $\blacksquare$
A counterexample is now straightforward. $X$ is compact by construction (and Hausdorff, so it is normal as well—it is even metrizable!), but the conjecture fails: $$X=\bigcup_{n\in\mathbb N\cup\{0\}}S_n,$$
and there is no finite subcover.
