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Munkres definition says the following:

A subbasis $S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $S$ is defined to be the collection $\mathcal T$ of all unions of finite intersections of elements of $S$.

Now, a collection of subsets of $X$ whose union equals $X$ could be made up of a collection of disjoint subsets of $X$. In that event, the collection $\mathcal T$ as defined above is an empty set? How does it generate the topology then?

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In this case, $T$ will consist of all unions of elements of $S$, along with $\emptyset$. Indeed, as you said, the intersection of any two or more elements of $S$ is $\emptyset$, but one element on its own counts as well for an intersection.

Indeed, given any family $\mathcal F$ of sets, the intersection of $\mathcal F$, noted $\bigcap\mathcal F$, is the set of elements that belong to all the elements of $\mathcal F$:

$$\bigcap\mathcal F:=\{x\,\mid\,\forall y\in\mathcal F,\,x\in y\}.$$

A finite intersection of elements of $S$ means $\bigcap\mathcal F$ for a finite $\mathcal F\subset S$. In particular, if $U\in S$, $\bigcap\{U\}=U$.

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One finite intersection of the sets is the original sets themselves...

Consider for instance the discrete topology $\tau$ on a set $X$. Then the singletons $\{x\}$ would be a subbasis... Every subset is then open, as a union of these singletons, which are intersections of a single set with itself (the trivial intersection, so to speak).

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The collection $T$ as defined in your question will not be empty.

If $S$ is a partition of $X$ then the collection of finite (non-empty) intersections of $S$ will be $S\cup\{\varnothing\}$. This collection serves as base of the topology and a set will belong to this topology if and only if it can be written as a union of elements of $S$.

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