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!

It is false.
Consier $\Bbb R$ with usual topology, $A=[0,2]$, $B=[1,3]$.
$$\partial(A\cup B)=\{0,3\}$$
$$\partial A\cup\partial B=\{0,1,2,3\}$$
Just think that some of the border of $A$ can be in the interior of $B$.
