$S$ is closed under pairwise unions $⇒$ $S$ is closed under arbitrary unions?

No, it's not the case. Let $S$ be the collection of finite subsets of $\mathbb{N}$, for example.

It is not the case. For example, the collection of (topologically) closed sets in $\Bbb R$ is closed under pairwise unions. However, the union $$ (0,1) = \bigcup_{n = 1}^\infty [1/n,1-1/n] $$ is not a (topologically) closed set.

You can also take $X=\mathbb{R}^2$ and let $S$ be the collection of bounded subsets. The union of two (or finitely many) bounded sets is bounded [prove it]. However, you can easily cover the entire plane with bounded "pieces" or "tiles" if you allow infinitely many of them [try that].