Definition of a compact set From the definition of a compact set, a set $A$ is said to be compact if for every open cover of $A$, there exists a finite subcover such that $A$ is a subset of this finite subcover.
Here if we consider the open cover, is it necessary for all the elements (which are sets) of this open cover to be of finite?
Also should the finite subcover have elements which are finite?
If we consider the subcover to have the set $(-\infty, +\infty)$ as one of its sets, then we have that all sets of $R$ compact.
But while proving theorems related to compact sets (from Rudin), this property of finiteness of the sets is never used. So should the elements of the finite open cover be finite or that  we should not include the set $(-\infty, +\infty)$ in specific. 
 A: No, finiteness of the individual sets is not requested nor is it useful.
$(-\infty,\infty)$ is one special cover of $\mathbb{R}$. The definition of compactness however demands that every open cover has a finite subcover. if you take, e.g., the cover of $\mathbb{R}$ consisting of the intervals $(n-1, n+1)$ with $n\in \mathbb{Z}$, then you will not find a finite subcover.
(The general definition of compact set you are citing takes some getting used to. For the beginning and for subsets of $\mathbb{R}$ or $\mathbb{R}^n$ the property which is easiest to grasp is that a set is compact if and only if it is bounded and closed).
A: As you say, a set is called compact if every open cover admits a finite subcover. More specificly, given any collection of open sets (no further restrictions) such that $A$ is contained in the union of them, you should be able to find finitely many of these open sets such that $A$ is still contained in the union of these finitely many open sets.
In case of the real numbers, $(-\infty, +\infty)$ is one (finite) open cover for any subset $A$ of $\mathbb{R}$, but in order to conclude that $A$ is compact, you should have that all open covers admit a finite subcover.
Example: the open interval $(0,1)$ can be covered by the intervals $(1/n,1)$, $n\in\mathbb{N}$, but this cover does not have a finite subcover. So although you can find finite open covers easily, the condition for compactness is in this case not satisfied.
