Any class of a nonempty $X$ generates a base for a topology on $X$. 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 $X \in \mathcal{B}$?
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 $B_3 \subseteq B_1 \cap B_2$.


See for your example if you do not include $X$ to the set $\mathcal{B}$,  it's still a basis.
A base for a topology is a set of subsets $\mathcal B\subset P (X)$ such that the set of all unions of elements of  $\mathcal B $ is the topology.
In general,  you are right that, as in this case, $X $ may not be included in the set of intersections of the elements of a subbase.  $X $ thus needs to be included,  in every event, to get a topology.