Why can't a open interval in $\mathbb{R}$ be compact? If I choose $(0,1)$ and do a covering like this:
$(-1,1/2)$ and $(1/3,2)$ it's a finite covering, then $(0,1)$ is a compact set?
 A: 
If I choose $(0,1)$ and do a covering like this: $(-1,\frac{1}{2})$ and $(\frac{1}{3},2)$ it's a finite covering, then $(0,1)$ is a compact set?

You are misunderstanding the definition of a compact set. The definition starts with "For every open coverings, there exists a sub coverings......". So, in order to show that $(0,1)$ is compact, you first have to prove something that is true for every open coverings of $(0,1)$, which is quite hard.
The example you gave $(-1,\frac{1}{2})\cup(\frac{1}{3},2)$ is indeed an open covering. However, this does not help you much since you will need every possible open coverings.
In fact, to show a set is not compact, all you need is just one counterexample which is already given in other answers. To show a set is compact, however, is much more difficult. Sometimes you can use the definition of compactness directly but as I said, doing something to every possible open covering sounds difficult. In these case, people usually either argue by contradiction or use the fact that being closed and bounded sometimes implies compactness.
A: $(0,1)$ is not compact.
Proof: Consider the open covering $\displaystyle\left\{\left ({1\over n}, 1 - {1\over n}\right) \right\}_{n=3}^\infty$; this is an open covering of $(0,1)$ but it has no finite subcovering.
A: Take the sequence $\large a_n = \frac{1}{n}$, which lives in (0,1).
it converges to a point, $0$, which is not in your set (0,1).
and so we say that your set is not sequentially compact.
(and so it is not compact.)
A: Yet another argument (which may be useful): we are used to slogan that "in $\mathbb{R}^n$ compact means closed and bounded" (from this you already see that $(0,1)$ not being closed cannot be compact) however compactness is quite different in nature from properties of being bounded and closed:
-being bounded is only with respect to some metric: it requires finer structure on our topological space
-closedness is not intrinsic property: the same set may be closed or not, depending on how you embedd it in some ambient space.
This is not the case for compactness!
-compactness does not require metric structure (only topological)
-compactness is intrinsic (independent from the embedding).
And why I'm talking about this? Since $(0,1)$ is topologically the same as $\mathbb{R}$ so if $(0,1)$ would be compact, so would be $\mathbb{R}$ and for the latter probably you won't have doubts that it is not. 
