Isn't basic open cover same as an open cover? I am reading topology by Gf Simmons and I came across a theorem on compactness which stated A topological space is compact if every open basic cover has a finite subcover.
But it defined a basic open cover as an open cover of $X$ whose sets are all in some given open base. But given $X$ and a topology defined on it a set consisting of all open sets of $X$ is always a basis (the largest there can be maybe) hence given any open cover its sets will always belong to this large basis. Doesn't it imply that every open cover is a basic open cover. So aren't those two necessarily the same?
 A: A more precise statement is the following: let $(X,\mathcal{T})$ be a topological space and let $\mathcal{B}$ be a base for the topology. Then $X$ is compact iff every cover by members of $\mathcal{B}$ has a finite subcover. 
This implies that we, to check compactness of a space $X$, “only” need to check a more limited set of open covers, namely the basic open covers.
The proof of the fact is not too hard (but does use the axiom of choice in general):
If $X$ is compact, of course a basic open cover has a finite subcover (these are just special open covers to which the definition of compactness applies), but for the converse let $\mathcal{O}$ be an arbitrary open cover of $X$. For each $x \in X$ pick some $O_x \in \mathcal{O}$ such that $x \in O_x$. As $\mathcal{B}$ is a base for the topology we can also pick a $B_x \in \mathcal{B}$ such that $$x \in B_x \subseteq O_x\text{.}$$
Then $\{B_x: x \in X\}$ is a basic open cover and so by assumption has a finite subcover, say $\{B_{x_1},\ldots,B_{x_N}\}$. But then it’s also clear that $\{O_{x_1},\ldots, O_{x_N}\}$ is a finite subcover of $\mathcal{O}$, as required.
A bit more involved is the Alexander subbase theorem, which is like the above proposition, but uses a subbase for the topology instead of a base. This allows us to give a relatively easy proof for Tychonoff’s theorem.
