Is this a valid proof that the boundary of a set on a metric space is closed? The definition that I have been given of the boundary of a subset $A$ of a metric space $X$ is:
$$\partial A=\{x\in X:\forall r\in \mathbb{R}, B_r(x)\cap A \neq\emptyset \text{ and } B_r(x)\cap A^c\neq\emptyset\}$$
So with this definition, we would have
$$(\partial A)^c=\{x\in X:\exists r\in \mathbb{R}, B_r(x)\cap A =\emptyset \text{ or } B_r(x)\cap A^c=\emptyset\}$$
$$(\partial A)^c=\{x\in X:\exists r\in \mathbb{R}, B_r(x)\cap A =\emptyset\}\cup\{x\in X:\exists r\in \mathbb{R}, B_r(x)\cap A^c=\emptyset\}$$
$$(\partial A)^c=\operatorname{int}A\cup \operatorname{int}A^c$$
Thus, the complement of the boundary of $A$ is the union of two open sets, in consequence, it is open. Then, $((\partial A)^c)^c=\partial A$ is closed.

Yes, that’s fine. You can also do it a little more directly: the definition of $\operatorname{bdry}A$ says that $x\in\operatorname{bdry}A$ iff $x\in\operatorname{cl}A\cap\operatorname{cl}(X\setminus A)$, i.e., that $\operatorname{bdry}A=\operatorname{cl}A\cap\operatorname{cl}(X\setminus A)$, and since this is the intersection of two closed sets, it is closed.
While your definition of the boundary is stated in terms of metric spaces, it’s just a special case of the definition for topological spaces in general, and the same argument shows that $\operatorname{bdry}A$ is closed in all topological spaces.
