Let $(X, d)$ be a metric space and let $A \subseteq X$ be a subspace, show that any open neighborhood of a boundary point of $A$ contains an interior point and an exterior point
The definition I'm working with for boundary point in this case is that $p \in X$ is a boundary point of $A$ if it is not an interior point and it is not an exterior point.
What I tried so far Let $p \in X$ be a boundary point of $A$. By definition $p$ is not an interior point nor is it an exterior point of $A$.
Let $U$ be an open neighborhood of $p$ in $X$ then by definition of $U$ being open we have $B(p, r) \subseteq U$ for some $r > 0$.
Suppose $B(p, r)$ contained no interior points of $A$. Then every $y \in B(p, r)$ has the property that for all $\delta > 0$ we have $B(y, \delta) \not\subseteq A$.
Hence for all $\delta > 0$, there exists a $z_{\delta} \in B(y, \delta)$ such that $z_{\delta} \not\in A$.
That was where I got stuck, what I was trying to do was argue by contradiction that $B(p, r)$ contained no interior points of $A$ and showing that we arrived at a contradiction and then arguing by contradiction that $B(p, r)$ contained no exterior points of $A$ and showing that we arrived at a contradiction, but I'm not sure how to do it by contradiction.
How could I go about proving this theorem?
The statement is false. If $\mathring A=\emptyset$ , that is, if $A$ has not interior points, then how can a neighborhood of a point of the boundary of $A$ contain an interior point of $A$? This happens, for instance, if $A=\{a\}$, where $a$ is a non-isolated point of $X$.