Let S ⊂ ℘(X). Let T be the coarsest topology on X which contains S. Then we call T the topology generated by S.
Let S ⊂ ℘(X). Then we can easily prove using (generalised distributive law) the topology generated by S is {union over a in A(intersection of D(a, j) for j in I(a)) where A is arbitary index set and I(a) is finite set for each D(a, j) on S}.
This fact used in proving very important result about basis: (B ⊂ ℘(X) is a basis if finite intersections of members of B are also unions of elements of B.) Now we can prove topology generated by a basis B consists of only those elements which are arbitary union of elements of B.
But in generalised distributive law we use cartesian product and for this to be non empty we have to AC. Hence using the fact about basis we use AC. Is there any other proof which don't use AC?
You're not making any choices, so you don't appeal to the axiom of choice. You just close $S$ under finite intersections, and then close this under unions. The axiom of choice is used, at best, to make some sort of cardinality arguments. But we don't care about that.
In here we don't really need to make any appeal to the axiom of choice regarding the distributivity, since the intersections are finite. Again, the simplest way to see this would be to see that you can first close under finite intersections (which we can do by induction, without appealing to choice) and then under unions (which is really just a single step).