This issue comes from An Introduction to Econometric Theory from Gallant (1997). On his way to describing a $\sigma$-algebra, Gallant says:
. . .let $\mathscr A$ denote the collection of the sets of the form (a,b] with 0 $\le$ a $\lt$ b $\le$ 1, finite unions of such sets, plus the empty set {}. . .
He claims that this is closed under compliment:
. . .Note that . . . (ii) ~A $\in$ $\mathscr A$ whenever A is . . .
However, this seems false. Let S = (0,1], which is clearly in $\mathscr A$. ~S, then, is [0], which isn't in $\mathscr A$. Moreover, Gallant suggests that one of the defining features of a $\sigma$-algebra that contains $\mathscr A$ (apparently in contrast with $\mathscr A$ itself) is that it admits intervals of the form (a,b), [a,b], and [a,b). But it seems like, if $\mathscr A$ is closed under compliment, it would have intervals of these forms, too. How am I confused?
Based on your quotes, I think $\mathscr A$ is supposed to be an algebra on $(0,1]$ rather than -- as you seem to assume -- on $[0,1]$.
Then your $S$ is not a problem, because the complement of $S$ is now $\varnothing$ which is explicitly in $\mathscr A$.
Being closed under complement does not make $\mathscr A$ contain intervals of the form $(a,b)$, $[a,b]$, or $[a,b)$.
Consider for example, $T=(\frac13,\frac23]$ Its complement is $(0,\frac13]\cup (\frac23,1]$. Note that both $T$ and its complement consists of intervals that are open to the left and closed to the right. Complements cannot make the left end be closed or the right end be open -- and neither can finite unions and intersections.
Once you extend $\mathscr A$ to a $\sigma$-algebra, you can also create countably infinite unions and intersections. These operations allow you to create, for example $(\frac13,\frac23)$ as the union $\bigcup_{n>4} (\frac13,\frac23-\frac1n]$.