Proof
(related to Proposition: Construction of Topological Spaces Using a Subbasis)
 By hypotheisis, $X$ is a set and $S$ is an arbitrary set of its subsets.
 We define $\mathcal O(S)$ as the set union of all possible finite intersections of elements of $S.$
 Obviously, $\mathcal O(S)$ is a topology:
 Verifying the first axiom: $\emptyset\in \mathcal O(S)$, since, by convention the empty union equals $\emptyset,$ and $X\in \mathcal O(S),$ since the empty intersection equals $X.$
 Verifying the second axiom: By construction, every finite intersection of elements of $S$ is in $\mathcal O(S).$
 Verifying the third axiom: By construction, every union of finite intersections of elements of $S$ is in $\mathcal O(S).$
 Moreover, by construction, $S$ plays exactly the role of a subbasis of $\mathcal O(S).$
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References
Bibliography
 Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970
 Jänich, Klaus: "Topologie", Springer, 2001, 7th Edition