$A\subset\Bbb R^n\text{ open}, p\in\Bbb R^n-A$. Then $A\cup\{p\}$ is open $\iff$ $p$ is isolated on $\partial A$.

Let $A\subset\Bbb R^n$ open and $p\in\Bbb R^n-A$, where $n>1$. Then $A\cup\{p\}$ is open if, and only if, $p$ is isolated on $\partial A$.

Tried a few things but none of them worked as expected, could you give a clue to tackle this problem ?

Edit (after a long time):

Initially my expectations were that $A$ looks like $\Bbb R^n$ minus countable points. But same ideia looks to be true if we omit countable points from a open set. Let me explain my new idea:

I was thinking if some reformulation of this problem would help. Then I found this one: Let $X\subset\Bbb R^n$, where $n>1$, be any set and take a point $x\in X$. Then, $x$ is an interior point of $X$ if, and only if, $x$ is isolated on the boundary of $X-\{x\}$. And this is essentially the fact that the interior of a set is disjointed from it's boundary.

Am I right ?