Help with case one of "the boundary of union is the union of boundaries when the sets have disjoint closures"

Question

Maybe I'm missing a key fact, but how does one go from $x \notin \overline{ A \cup B } = \overline{A} \cup \overline{B}$ to $x \notin \partial A \cup \partial B$ in the first case of this answer? Should I use $\bar A \cup \bar B = \left( {A \cup \partial A} \right) \cup \left( {B \cup \partial B} \right) = A \cup B \cup \partial A \cup \partial B$ in some way?

Answer 1

It's just because $\overline{A} = A \cup \partial A$, so if $x\notin \overline{A}$ (i.e., $x\notin A \cup \partial A$), then certainly $x\notin \partial A$.

Combining this with the same statement for $B$, you obtain the desired result.

Said more concisely, it's because $\partial A \subset \overline{A}$.