I'll take almost your example :
$$U = \bigcup_{n\in\mathbb N^*} (\frac1n, 1)$$
cover (0,1) and is an "open cover".
Let's take only a finite set of $\{ (\frac1n, 1) | n\in\mathbb N^*\}$, say $\{(\frac 1n,1)| 0< n <100\}$. (100 is purely arbitrary, in order to make things the more concrete that I can).
If you consider: $$U_{100} = \bigcup_{n=1}^{100} (\frac1n, 1),$$ the number $\frac1{101}$ is not in $U_{100}$.
And whatever the number of subsets you're taking, you'll be able to find a number which is not in you subcover (do you want the formal proof of that ?).
EDIT (proof of that):
Let $C$ be a finite subset of $E=\{ (\frac1n, 1) | n\in\mathbb N^*\}$ (note that it's possible to have "holes", for example $\{(\frac1{100},1), (\frac12,1)\}$ is a finite subset of $E$).
Let $N$ the greater integer such that $(\frac1{N},1)\in C$ (there is effectively a "greater integer such that" beacause $C$ is finite). The number $\frac1{N+1}$ does not belong to the union of the elements of $C$.
Thus we've proved that whatever the finite subset of E we choose, this subset doesn't cover $(0,1)$.
END OF EDIT
For your other question, @user2345215 pointed out that a "thing" can be itself infinite, so there'is no contradiction to be infinite and to be cover by a finite number of things (each thing can be infinite).
Best Answer
An open cover of $(-2,2)$ can only be broken down into $(-2,0) \cup (0,2)$ if those two open sets are included in the cover, not to mention that $(-2,0) \cup (0,2)$ fails to cover $(-2,2)$ to begin with (what contains $0$?).
You are right when you say that open sets of $\mathbb R$ are not compact (well the empty set is). To show this, you would have to construct a open cover which can never be reduced to a finite subcover. The simplest open set to look at for this would be $(0,1)$.
To address your main question, is a singleton, say $\{0\}$, compact? What about a set of two real numbers? Three? Some natural number $n$?