Let $\overline{A} = \{x : x \text{ is a boundary point of A}\}$ and $\mathring{A} = \{x : x \text{ is a interior point of A}\}$ we will show that $\overline{A}$ is closed and $\mathring{A}$ is open.
In order to show that $\mathring{A}$ is open we have to show that every element of $\mathring{A}$ is an interior point. Since any $x\in\mathring{A}$ is an interior point of $A$ we now that there exist an $r>0$ s.t. the open ball $B(x,r)\subseteq A$. But since $B(x,r)$ is open we now that for any $x^*$ we can find $r^*>0$ s.t. $B(x^*, r^*) \subseteq B(x,r) \subseteq A$. But this shows that every element of $B(x,r)$ is an interior point of A and thus the open ball $B(x,r)$ is also contained in $\mathring{A}$, i.e. $B(x,r) \subseteq \mathring{A}$. Thus, every element of $\mathring{A}$ is an interior point and we conclude that $\mathring{A}$ is open.
In order to show that $\overline{A}$ is closed we have to show that $\overline{A}$ contain all its boundary points. In order for this to happen it must be the case that $\overline{A}$ and $A$ share the same boundary points.
Thus, let $x_0$ be a boundary point of $\overline{A}$, we will show that $x_0$ is also a boundary point of $A$. Since $x_0$ is a boundary point of $\overline{A}$ we now that for every $r>0$ we can find $x,y\in B(x_0, r)$ s.t. $x\in\overline{A}$ and $y\in \overline{A}^\mathrm{C}$. We know that $x$ is a boundary point of $A$ and thus for every $r^*>0$ we can fins $x^*,y^*\in B(x, r^*)$ s.t. $x^*\in{A}$ and $y^*\in{A}^\mathrm{C}$. If we can show that their exist an $r^*>0$ s.t. $B(x,r^*) \subseteq B(x_0, r)$ then we now that $x^*,y^* \in B(x_0, r)$ where $x^*\in{A}$ and $y^*\in{A}^\mathrm{C}$ we have shown that $x_0$ is a boundary point of $A$ and thus $\overline{A}$ contain all its boundary points and is closed.
Set $r^* = r - d(x, x_0)$ it then follows that $d(y^*,x_0)\leq d(y^*,x) + d(x, x_0)< r-d(x, x_0) + d(x,x_0) = r$ which shows that $B(x,r^*) \subseteq B(x_0, r)$ and hence $B(x_0, r)$ contains at least one $x^*\in A$ and $y^*\in A ^\mathrm{C}$ $\implies$ $x_0$ is a boundary point of A $\implies$ $x_0\in{\overline{A}}$ $\implies$ $\overline{A}$ is closed.
