Why is there no finite subcover for open sets? If $A$ is an open and bounded set, then why is $A$ itself not a finite subcover of $A$?
 A: Given an open set $A$, the collection $\{A\}$ is an open cover of $A$. However, the criterion for compactness, which is what I assume you are concerned with, requires that every open cover of $A$ has a finite subcover. Therefore, it is not enough to claim that just because one open cover, $\{A\}$ of $A$ has a finite subcover, then every open cover of $A$ does.
A: Compactness is not about finding an open cover in general. Every topological space has an open cover (namely, the whole space itself). Compactness is a game that is being played. If I give you an open cover, can you find a finite subcover? For instance, let's consider $(0,1)$. If I give you $\{(0,\frac{1}{2}), (\frac{1}{3},1)\}$ as a cover of this space, you can clearly give me a finite subcover.
On the other hand, if I give you $\bigcup_{n\in \mathbb{N}} \{(\frac{1}{n},1)\}$ as a cover of this space, there is no way you can find a finite subcover. Therefore, this space is not compact (under the usual topology given to this space). 
A: I think the issue here is that you are assuming that if someone gives you a cover $\{U_{\alpha}\}$ for a bounded open set $A$ then it is the case that one of the $U_{\alpha} = A$ which is not true. Consider this example,
