Is a subbase an open cover? 
Definition
A subbase for a topology $\cal T$ on a set $X$ is a subcollection $\cal S$ of $\cal T$ such that the collection
$$
\mathcal I:=\Big\{Y\in\mathcal P(X):Y=\bigcap\mathcal S'\,\text{with $\mathcal S'\subseteq\mathcal S$ finite}\Big\}
$$
is a base for $\cal T$.

So I am trying to understand if a subbase is always an open cover when $X$ is not empty. So I observe that if $X$ is not empty then for any $x\in X$ there exists $\cal S'\subseteq S$ with $n\in\omega$ such that
$$
x\in\bigcap \cal S'
$$
but (clik here for details) we know that $\bigcap\emptyset$ is not defined in ZF so that actually there exists $S_1,\dots, S_n\in\cal S$ not empty with $n\in\omega$ such that
$$
x\in S_1\cap\dots\cap S_n\subseteq\bigcup\cal S
$$
which proves that $\cal S$ cover $X$.
Now here Mirko and even here Henno state on the contrary that a subbase is not an open cover giving as counterexample the empty collection with respect the trivial topology into a not empty set $X$: however if $\emptyset$ was a subbase for $\{\emptyset,X\}$ then there must exist $x\in X$ such that $x$ lies in $\bigcap\cal F$ for any $\cal F\subseteq\emptyset$ but in this case $\cal F$ must be empty and $\bigcap\emptyset$ is not defined so that I argue that a subbase is not a cover only if we informally put
$$
X:=\bigcap\emptyset
$$
for any not empty set $X$, right?
So I ask to explain well if a subbase is an open cover with respect the above definition and then I ask if a cover is a subbase for any topology. So could someone help me, please?
 A: There are two ways to define the notion of a subbase for a topology.
The most-used way is to say $\mathcal{S}$ is a subbase for the topology $\tau$ if the family of intersections of finite subfamilies of $\mathcal{S}$ is a base for $\tau$. The way the latter sentence is explained further generally leads to the tacit assumption that $\mathcal{S}$ covers the underlying set $X$. For example, in Engelking's General Topology the finite intersections are expressed as $S_1\cap\cdots\cap S_k$, so one is lead to believe that only non-empty finite subfamilies are being used. In Kelley's General Topology a topology is just a family of sets and the underlying set is derived from it; in chapter 1, Theorem 12, the family of finite intersections of an arbitrary family of sets, $\mathcal{S}$, is shown to be a base for a topology on $X=\bigcup\mathcal{S}$. In these contexts a subbase is indeed a cover.
Some people call $\mathcal{S}$ a subbase for $\tau$ if $\tau$ is the smallest topology that contains $\mathcal{S}$. In that case the family $\mathcal{S}=\bigl\{\{0\}\bigr\}$ is a subbase for the Sierpinski topology $\tau=\bigl\{\emptyset,\{0\},\{0,1\}\bigr\}$ on the set $X=\{0,1\}$, even though it is not a cover of $X$. In this context the corresponding base for the topology consists of all finite intersections as above, with the whole set $X$ added to it as a member, either explicitly or by stipulating $X=\bigcap\emptyset$ as explained in this answer. In this context a subbase need not be a cover.
When I teach Topology I explain both ways and warn the students to check the definition of the book that they are reading and stick to that. Note that the first definition implies the second and that the second implies the first if the family is a cover, so for covers there is no ambiguity. Thus every cover serves as a subbase for some topology.
A: A subbase doesn't have to be an open cover at all.
The point is that any subset $S\subset \mathscr P(X)$ of the power set of $X$, that's any collection of subsets of $X$, generates a topology,  the smallest topology (or the intersection of all topologies) containing $S$.
So to reiterate,  any set of subsets is a subbase (for the topology it generates).
A basis for a topology,  otoh, has to be a cover. To get from a subbase to a base, take all finite intersections of elements of $S$, together with the set $X$.
