To be compact, every open cover of the set must have a finite subcover. If you can find any open cover that does not have a finite subcover, then the set is not compact.
The covers need not be stated as a sequence (they could, in fact, be uncountable). However, in $\mathbb R$, the union of any collection of open sets is equal to the union of a countable subset of that collection. So it's sufficient to use sequences of open sets for $\mathbb R$.
It's also important to have an intuition for why the cover must be open. Nonempty open sets necessarily have nonempty interior. In $\mathbb R$, this means they have nonzero extent. Additionally, to cover any set with nonempty interior, the open sets inevitably must have a nonempty overlap, which is itself an open set and has nonzero extent. To handle the endpoints of a closed set, the open sets also overhang the edges. This is what forces the existence of a finite subcover for closed, bounded sets in $\mathbb R$--the overlaps cut off any sequence of open intervals that approaches a point but never reaches it.