explanation for why is the following set not compact The set A$= (0,1] \subset{R}$ equipped with the usual topology, is not a compact set because it's not closed and bounded. Going by the choice of covers, we can choose an open set U $= (\frac{1}{i},2)$ and say that we can not choose finite sub-covers of the type-U such that A $\subset \bigcup_i^n U_i.$ I understand this part well, but couldn't we choose two open sets such as B$= (-1,2)$ and C$= (-2, 2)$ and say that A $\subset (B \cup C)$ and hence I have two sub-covers(finite) which cover the original set A and hence A is compact? 
Where is my argument's flaw?
 A: To prove something is compact, you have to prove that any open cover has a finite subcover. 
You have just shown a particular cover that has a finite subcover, which does not prove that it is compact.
A: Here's an analogy:
Definition: A prime number is a positive integer greater than 1 whose only positive whole factors are itself and one.
Counter Example:  15 is not prime because it has factors 3 and 5.
Confused student:  But what if we chose factors 1 and 15.  Those would be factors that are only itself and one.  So it should be prime.
... Now lets do the same with compact...
Definition:  A compact set is a set whose only open covers will have finite subcovers.
Counter example:  $(0, 1]$ as it has open cover $\{(1/i, 2)\}$ that has no finite subcover.
Confused student:  But what if we chose the open cover $\{(0,2)\}$.  It is finite cover.  Why isn't it compact?
.....
Notice:  Every set, open, closed, bounded, unbounded, compact, not compact, EVERY set has a finite open subcover.  Just take $\mathcal U = X$ where $X$ is the entire space.  That is an open cover.  And it's as finite as it gets.
