If $S\in\left(NP\bigcup coNP\right)$ then $\overline{S}\in NP\bigcap coNP$?

Is it true that if $$S\in\left(NP\bigcup coNP\right)$$ then $$\overline{S}\in NP\bigcap coNP$$?

I couldn't find any answer to that question.

My attempt at proving it:

If $$S\in\left(NP\bigcup coNP\right)$$, then, by set theory, $$S=A\bigcup B$$ for some $$A\in NP\wedge B\in coNP$$.

Therefore the complement of $$S$$ is $$\overline{S}=\overline{A\bigcup B}=\overline{A}\bigcap\overline{B}\in NP\bigcap coNP$$

There is a distinction between $$S\in (NP\cup co-NP)$$, and $$S\in (NP\lor co-NP):=\{A\cup B|A\in NP, B\in co-NP\}$$
You assumed both are the same, however the first talks about $$S$$ being in either $$NP$$ or $$co-NP$$, while the second talks about $$S$$ being a union of something from $$NP$$ and something else from $$co-NP$$.
About the question in title, we actually don't know. If $$NP\neq co-NP$$, then there is some $$L\in NP\setminus co-NP$$ (or the other way around), which means that $$L\in (NP\cup co-NP)$$ but $$\overline{L}\notin (NP\cap co-NP)$$ (since $$L\notin NP$$ then $$\overline{L}\notin co-NP$$)
But if $$NP=co-NP$$, then clearly what you stated would be trivially true.
So essentially, what you tried to prove is equivalent to the open problem "Is $$NP=co-NP$$ or not?".