Clarification on finite intersection I am trying to understand the following example

Let $X=\{1,2,3 \}$ and $\mathscr{S}=\{ \{1,2\},\{2,3\}  \}$ so that $\cup_{S\in\mathscr{S}}S=X$. If we take the collection $\mathscr{B}$ of all sets that can be obtained by intersecting a finite number of members of $\mathscr{S}$ we have $\mathscr{B}=\mathscr{S}\cup\{X,\{2\}  \}$.

Now I can see why $\mathscr{S}$ in $\mathscr{B}$ as $\{1,2\}\cap\{1,2\}=\{1,2\}$  and $\{2,3\}\cap\{2,3\}=\{2,3\}$. Similarly, $\{1,2\}\cap\{2,3\}=\{2\}$, but how do we get $X$ in $\mathscr{B}$? 
EDIT: I think there is a bit of confusion. To set things correct $\mathscr{S}$ is a collection of subsets of $X$ such that $\cup S=X$. This example is given in Stromberg Introduction to Classical Analysis, p.96 (example f).
 A: There is no way to get  $ X$ in $\mathscr{B}$ because we can not describe $X$ as an intersection of members of  $\mathscr{S}$
This $$\mathscr{B}=\mathscr{S}\cup\{X,\{2\}  \}$$
is simply not true unless we change the usual definition of intersection to include $X$ an mentioned in Arnaud Mortier 's solution..
A: Here is a way for this to make sense.
Define the intersection of a family $\{A_i\}_{i\in I}$ of subsets of $X$ as $$\{x\in X\ | \ \forall i\in I, x\in A_i\}$$
Then it is a classical result in logic that any proposition that starts with "$\forall i\in \emptyset$" is TRUE. Because to prove it wrong you would have to find some $i$ in the empty set such that the conclusion does not hold. No matter what the conclusion of the proposition is, you can't find an $i$ in the empty set.
Therefore, when $I=\emptyset$, all elements of $X$ match the condition and therefore you get $X$ as the intersection of an empty family. 
Note that this relies on a definition of intersection of subsets of $X$, rather than the usual definition of intersection of sets in an absolute sense.
A: We compute from $\mathcal{S}$ all intersections of its finite subfamilies:
The empty subfamily $\emptyset \subseteq \mathcal{S}$ has intersection $\bigcap \emptyset = X$ by common convention (as $\forall_{y \in \emptyset} : x \in y$ is true for all $x \in X$).
All subfamilies with one member $\{S\} \subseteq \mathcal{S}$ just have that member as their intersection, so we get $\cap \{S\} = S$, for all $S \in \mathcal{S}$. 
and $\bigcap\{\{1,2\}, \{2,3\}\} = \{2\}$ so $\{2\}$ is also the result of such a finite intersection.
Hence we get $\mathcal{S} \cup \{X ,\{2\}\}$.
