We shall show that if $X$ is a $T_1$ space, then it is countably compact if and only if every infinite open cover has a proper subcover. The key idea is to proceed by contrapositive.
($\Rightarrow$ Needs $T_1$) Suppose that $ \ \mathcal{U} \ $ is an infinite open cover of $X$ with no proper subcover. Then for each $U \in \mathcal{U}$, there is a point $p_U \in U$ that doesn't belong to any other member of $ \ \mathcal{U} \ $. The set $A = \{p_U : U \in \mathcal{U} \}$ is infinite and doesn't have a limit point. Indeed, if $x \in X$, then there is $U \in \mathcal{U}$ containing $x$ and $ U \cap A = \{ p_U \}$. If $x = p_U$, then it isn't a limit point of $A$. If $x \ne p_U$, then, using $T_1$, $U- \{ p_U \}$ is open and $(U- \{ p_U \}) \cap A =\emptyset$. So $x$ isn't a limit point of $A$.
George Lowther gave an example showing that the $T_1$ hypothesis is fundamental. Since there are many comments, I'll reproduce it here:
"Consider the example of the real numbers where the open sets are unions of intervals $[n,a)$ for integer $n$ and real $a>n$. This is $T_0$ but not $T_1$. It is also countably compact, but the infinite cover $\mathcal{U}=\{[n,n+1)\colon n\in\mathbb{Z}\}$ has no proper subcovers. So, $T_1$ is needed."
A direct approach to prove this implication was suggested by Carl Mummert. Let $ \ \mathcal{U} \ $ be an infinite open cover of $X$ and $ \ \mathcal{U}_0 $ be a countably infinite subset of $ \ \mathcal{U} $. Now consider $ \ \mathcal{U}_1 = \mathcal{U} - \mathcal{U}_0$ and let $V \ $ be the union of all sets in $ \ \mathcal{U}_1$. Then $ \ \mathcal{U}_0 \cup \{ V \ \}$ is a countable open cover of $X$ and follows from $(1)$ that it has a finite subcover $U_1, \ldots, U_n$. Now the set consisting of all $U_i$, with possible exception of $V$, adjoined with the sets $ U \in \mathcal{U}_1$ is a proper subcover of $ \ \mathcal{U}$.
($\Leftarrow$ Doesn't need $T_1$) Now, suppose that $X$ isn't countably compact. Then there is an infinite set $A$ with no limit points. Thus $A$ is closed. By the definition of limit point, for each $x \in A$, there is an open set $U_x$ containing $x$ such that $ U_x \cap A = \{ x \} $. Consider $ \ \mathcal{U} \ $ the set of all $U_x$ plus, if necessary, $X-A$. Then $ \ \mathcal{U} \ $ is an infinite cover of $X$ with no proper subcovers.
P.S. Thanks Carl Mummert, George Lowther and Mark for your efforts to clarify and improve this answer.
One problem with this approach is that the union of $\{O_M\}$ with the finite subset $S$ that is associated with $x_n$ now covers the point $x_n$ but that does not imply that it covers all of $K$ as there may be much more of $K$ that is not covered by $S$ nor by $O_M.$ Another problem is the unwarranted claim that x is excluded from any finite sub-cover. This does not follow from the fact that x is the limit of "my sequence."
The def'n of compact is that $K$ is compact iff every open cover of K has a finite sub-cover. It is a theorem that a subset of the reals is compact iff it is closed and bounded.
Best Answer
Here we need to distinguish two quantities that we are counting:
Now certainly as you have suggested, every open set in $\mathbb{R}$ can be decomposed into countably many open intervals. This is related to the second quantity in the above.
However, the word finite in the definition of compact refers to the first quantity in the above. In other words, by finite subcover, we mean that the open sets are chosen from the original open cover and the number of such sets are finite.
So whether the open sets can be further decomposed into open intervals is irrelevant in the definition of compactness.
Edit: In response to the OP's follow up question, I would like to use a rather trivial example to emphasize that it is possible for some open set in a particular finite open cover to be a disjoint union of infinitely many open intervals. Consider the set $ [0,1]$ in $ \mathbb{R}$. This set is compact. Now consider this specific open cover $$ \mathcal{O} = \lbrace (-1,2), [0,1]-\mathcal{C} \rbrace, $$ where $\mathcal{C}$ is the Cantor set. There are two open sets in $ \mathcal{O} $, so $ \mathcal{O} $ is a finite open cover of $ [0.1] $. But the set $ [0,1] -\mathcal{C} $ is a disjoint union of countably many open intervals. In particular, the number of open intervals is not finite.
In addition, if we write out $$ [0,1] - \mathcal{C} = \bigcup_i I_i $$ explicitly, where each of the $I_i$ is an open interval. Then actually each of the $I_i$ is not an element of the open cover $\mathcal{O}$.