Any class of a nonempty $X$ generates a base for a topology on $X$.

general-topology

Theorem Any class $\mathcal{A}$ of subsets of a nonempty set $X$ is the subbase for a unique topology $\tau$ on $X$. That is to say, the finite intersections of members of $\mathcal{A}$ form a base for the topology $\tau$ on $X$.

Example Consider $X=\{a,b,c,d\}$

Let $\mathcal{A}=\{\{a,b\},\{b,c\},\{d\}\}$

The finite intersection of members of $\mathcal{A}$ give us:

$$\mathcal{B}=\{\{a,b\},\{b,c\},\{d\},\{b\},\varnothing,X\}$$

How is it true that $X \in \mathcal{B}$ ? By definition of subbase, the finite intersection of members of $\mathcal{A}$ are the members of some base $\mathcal{B}$.

But for $a_i \in \mathcal{A}, \displaystyle\bigcap a_i \neq X$ for any finite intersection of $a_i$'s.

Look: $\{a,b\}\cap\{b,c\}\cap \{d\} \neq \{a,b,c,d\}$ nor do any other intersection of elements of $\mathcal{B}$ result in $X$. Anybody know what's going on here? Why is $X \in \mathcal{B}$?

The text you quoted seems too use the quite common “convention” of the empty intersection: an intersection of an empty family is the “universal set” in the context at hand.

So if we have $\mathcal{F} \subseteq \mathscr{P}(X)$, so a family of subsets, then $$\bigcup \mathcal{F} = \{x \in X\mid \exists F \in \mathcal{F}: x \in F\}$$ is a pretty uncontroversial definition for the union of the family $\mathcal{F}$. We can define analogously $$\bigcap \mathcal{F} = \{x \in X\mid \forall F \in \mathcal{F}: x \in F\}$$ and then the usual de Morgan laws like $$\left(\bigcup \mathcal{F}\right)^\complement = \bigcap \{F^\complement\mid F \in \mathcal{F}\}; \left(\bigcap \mathcal{F}\right)^\complement = \bigcup\{F^\complement\mid F \in \mathcal{F}\}$$ hold by simple first order logic, as well as simple facts like $$\bigcap \mathcal{F} \cap \bigcap \mathcal{G}= \bigcap (\mathcal{F} \cup \mathcal{G})\tag{1}$$ etc.

But the case where $$\bigcup \emptyset = \emptyset; \bigcap \emptyset = X$$ (the second by “void truth”) which is OK with de Morgan laws (as $X$ and $\emptyset$ are indeed each other’s complement) and also with a law like $(1)$ (as $X$ has no effect on the intersection). It’s also logical when seeing intersection as the infimum in the lattice $\mathscr{P}(X)$ (any member is a lower bound for the empty set, and so the largest lower bound is $X$, the maximal element of the lattice). And it’s nicely symmetric with the sup of the emptyset being the minimal element $\emptyset$. So in this context of the lattice of subsets of $\mathscr{P}(X)$, as in topology, it’s quite common practice to state that $\bigcap \emptyset = X$ and so $X$ is the intersection of a finite (size $0$) subfamily of sets of $X$. The empty union convention is already used a lot to have $\emptyset$ in the topology generated by any base anyway (the set of unions of subfamilies of the base is then nicely a topology). In this light the axiom $\emptyset, X \in \mathcal{T}$ for a topology are superfluous as they follow from the arbitrary union and finite intersection axioms (but it’s usually stated anyway).

My own initial general topology course did spend a little time the state this empty intersection “convention” (it makes no sense in general set theory as there is no universal set (Russell’s paradox etc.), but completely valid in the lattice of subsets, so that indeed any subfamily of $\mathscr{P}(X)$ forms a subbase (i.e. a generating family for a topology): the finite intersections include the subbase itself (as $\bigcap\{A\} = A$ etc.), $X$ for the empty intersection, and $(1)$ above shows this family is closed under finite intersections and so fulfills the conditions to be a base, etc.

You may or may not include $X$ to a basis. $X$ will have to be included by hand as one cannot get $X$ by taking intersections of proper subsets of $X$. Otherwise you can also include $X$ to $\mathcal{A}$ to start with.

The definition of basis is

For each $x \in X$, there is at least one basis element $B$ containing $x$.
If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subseteq B_1 \cap B_2$.