If you define $\tau(\mathscr{S})$ to be the intersection of all topologies $\tau$ on $X$ such that $\mathscr{S}\subseteq\tau$, then it’s trivial that $\tau(\mathscr{S})$ is the smallest topology on $X$ containing $X$: by definition if $\tau$ is any topology on $X$ with $\mathscr{S}\subseteq\tau$, then $\tau(\mathscr{S})\subseteq\tau$.
What’s not quite trivial is to prove that if $$\mathscr{B}=\left\{\bigcap\mathscr{F}:\mathscr{F}\subseteq\mathscr{S}\text{ is finite}\right\}$$ and $$\tau(\mathscr{B})=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;,$$
then $\tau(\mathscr{B})=\tau(\mathscr{S})$, i.e., that the process of closing $\mathscr{S}$ under finite intersections and then under arbitrary unions yields the smallest topology containing $\mathscr{S}$. (This is the other common definition of the topology generated by an arbitrary $\mathscr{S}\subseteq\wp(X)$.)
Added: I’ll show that $\tau(\mathscr{B})=\tau(\mathscr{S})$. The first step is to show that $\mathscr{S}\subseteq\tau(\mathscr{B})$.
$\mathscr{B}$ is the set of all intersections of finite subsets of $\mathscr{S}$, so in particular $\mathscr{S}\subseteq\mathscr{B}$. And $\tau(\mathscr{B})$ is the set of all unions of subsets of $\mathscr{B}$, so in particular $\mathscr{B}\subseteq\tau(\mathscr{B})$. Thus, $\mathscr{S}\subseteq\mathscr{B}\subseteq\tau(\mathscr{B})$, i.e., $\mathscr{S}\subseteq\tau(\mathscr{B})$.
The next step is to show that $\tau(\mathscr{B})$ is actually a topology on $X$.
By definition each $U\in\tau(\mathscr{B})$ is a union of members of $\mathscr{B}$, and a union of unions of members of $\mathscr{B}$ is itself a union of members of $\mathscr{B}$ and therefore a member of $\tau(\mathscr{B})$, so $\tau(\mathscr{B})$ is closed under taking arbitrary unions.
Suppose that $U,V\in\tau(\mathscr{B})$; then there are $\mathscr{U}\subseteq\mathscr{B}$ and $\mathscr{V}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$ and $V=\bigcup\mathscr{V}$. (That is, $U$ and $V$ are both unions of members of $\mathscr{B}$.) Then $$\begin{align*}U\cap V&=\left(\bigcup\mathscr{U}\right)\cap\left(\bigcup\mathscr{V}\right)\\&=\bigcup_{G\in\mathscr{U}}\left(G\cap\bigcup\mathscr{V}\right)\\&=\bigcup_{G\in\mathscr{U}}\bigcup_{H\in\mathscr{V}}(G\cap H)\tag{1}\;,\end{align*}$$ where each of the sets $G\cap H$ is the intersection of two members of $\mathscr{B}$. Each member of $\mathscr{B}$ is an intersection of finitely many members of $\mathscr{S}$; if you intersect two such sets, you still have an intersection of finitely many members of $\mathscr{S}$ and hence a member of $\mathscr{B}$. That is, each of the sets $G\cap H$ appearing in the big union in $(1)$ is a member of $\mathscr{B}$, so by definition their union, $U\cap V$, is a member of $\tau(\mathscr{B})$.
The only thing that remains to be checked is that $\varnothing,X\in\tau(\mathscr{B})$. Certainly $\varnothing\subseteq\mathscr{B}$, so $\bigcup\varnothing\in\tau(\mathscr{B})$ by definition, and $\bigcup\varnothing=\varnothing$. In words, $\varnothing$ is the union of the empty subfamily of $\mathscr{B}$. Finally, $\varnothing$ is a finite subfamily of $\mathscr{S}$, and it’s vacuously true that $$\bigcap\varnothing=\{x\in X:x\in S\text{ for all }S\in\varnothing\}=X\;.$$
We now know that $\tau(\mathscr{B})$ is a topology on $X$ with $\mathscr{S}\subseteq\tau(\mathscr{B})$, so by the first part $\tau(\mathscr{S})\subseteq\tau(\mathscr{B})$. On the other hand, if $\tau$ is a topology on $X$ such that $\mathscr{S}\subseteq\tau$, then it must be true that $\mathscr{B}\subseteq\tau$, because $\tau$ is closed under taking finite intersections. And then it must also be true that $\tau(\mathscr{B})\subseteq\tau$, since $\tau$ is closed under taking arbitrary unions. Thus,
$$\tau(\mathscr{S})\subseteq\tau(\mathscr{B})\subseteq\bigcap\left\{\tau:\tau\text{ is a topology on }X\text{ and }\mathscr{S}\subseteq\tau\right\}=\tau(\mathscr{S})\;,$$
and it follows immediately that $\tau(\mathscr{B})=\tau(\mathscr{S})$.
