Help proving $\partial (A \cup B) = \partial A\cup\partial B$? I know this is a duplicate but the other two haven't helped me much. 

Fist attempt: Tried proving through double inclusion, but wasn't sure of how to convey being an element of one implied being an element of the other in either direction, although I suspect from left to right would be the easier of the two.
Second attempt: Tried proving equality directly, using the fact that the boundary of a set is equal to the set difference of its closure and interior, but struggled proving closure of union is a union of closures, or the interior of a union is a union of interiors. 
I'm getting pretty frustrated and any help would be greatly appreciated!

Your intuition is fine with respect to the left to right inclusion:
$$ \partial(A\cup B)\subseteq \partial(A)\cup \partial(B) .$$
See a complete formal proof in this MSE link.
If $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, then the other inclusion holds too. See a proof in this other MSE link.
