Is it true that the boundary of an open ball is equal to the boundary of a closed ball, in an arbitrary metric space?

Question

Is my reasoning correct here?

Let $B(x;r)=\{y\in X : d(x,y) < r\}$, $\partial B(x;r)=\{y\in X : d(x,y) = r\}$, $\overline B(x;r)=\{y\in X : d(x,y) \leq r\}$, and $\partial\overline B(x;r)=\{y\in X : d(x,y) = r\}$.

Therefore $$\partial B(x;r)=\{y\in X:d(x,y)=r\}=\partial\overline B(x;r),$$ and the statement is true for every metric space. Additionally, we can consider an open interval $(a,b)=A$ in the real line $\Bbb R$. Then $\partial A=\{a,b\}$, $\overline A=[A,B]$, and $\partial\overline A=\{a,b\}$. Therefore $\partial A=\partial\overline{A}$.

Answer 1

No, $\partial B(x;r)$ is not necessarily $\{y \in X: d(x,y) = r\}$. E.g. let $X$ be the integers with the metric $d(x,y) = |x-y|$. For any integer $x$, $B(x; 1) = \{x\}$ and so is its closure; $\partial B(x,1) = \emptyset$. But $\{y: d(x,y) \le 1\} = \{x-1,x,x+1\}$.

Answer 2

Not valid. Examples:

(i). $X=\{x,z\}$ with $d(x,z)=1.$ Then $B(x,1)=\bar B(x,1)=\{x\}$ is open-and-closed so it has empty boundary. But $\{y\in X: d(x,y)=1\}=\{z\}.$

(ii). $X=[-1,1]$ with $d(u,v)=|u-v|.$ Then $\partial B(0,1)=\{-1,1\}$ but $\bar B(0,1)=X$ has empty boundary.