Rewriting the boundary, $\partial A$, of $A \subseteq (M,d)$. If we have an arbitrary metric space $(M,d)$ with $A \subseteq (M,d)$ then the boundary of $A$ is defined the following way for any $\epsilon > 0$
$\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$.
However, I understand we can rewrite this definition: let $int(A)$ be the interior of $A$ and $cl(A)$ be the closure of $A$. Then we can rewrite $\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$ as $$\partial A := cl(A) \cap cl(A^{c}).$$
While most of my colleagues in my program are just taking this at face value, I wanted to see why this is true; here is my reasoning

Since $\partial A :=$ $\{x \in M: B_{\epsilon}(x) \cap A \neq \emptyset, B_{\epsilon}(x) \cap A^{c} \neq \emptyset\}$, so $x \in \partial A$ if and only if no ball $B_{\epsilon}(x)$ is subset of either $A$ or $A^{c}$, which is true if and only if $x \notin int(A)$ and $x \notin int(A^{c})$. Therefore, $x \in (int(A))^{c}$ and $x \in (int(A^{c}))^{c}$. So we can write
$$\partial A := (int(A))^{c} \cap (int(A^{c}))^{c}.$$ Now, clearly this can be simplified - But this is where I start to have questions. Is it true that $int(A) = (cl(A))^{c}$? and if so, wouldn't that imply that $(int(A))^{c} = cl(A)$? If this is the case, then we could arrive at the following result by applying the same logic to ($int(A^{c}))^{c}$:
$$\partial A := cl(A) \cap cl(A^{c}).$$ Please correct me if I am wrong or if I am thinking about something backwards.
 A: Your definition of the boundary is slightly incorrect it should be:
$$
\partial A := \{x \in M : \mbox{for all}~\epsilon > 0, B_{\epsilon}(x) \cap A \neq 0, B_{\epsilon}(x) \cap A^c \neq 0\}.
$$
In other words, the quantifier on $\epsilon$ should be universal.

To see the equivalent characterization, namely that $\partial A = \mathrm{cl}~A \cap \mathrm{cl}~A^c$, we as usual check the two inclusions.
First inclusion: Let $\epsilon > 0$. Suppose that $x \in \mathrm{cl}~A$ and $x \in \mathrm{cl}~A^c$. Then $B_{\epsilon}(x) \cap A$ and $B_{\epsilon}(x) \cap A^c$ are nonempty, and therefore:
$$
\partial A \supset \mathrm{cl}~A ~\cap\mathrm{cl}~A^c.
$$
Second inclusion: Conversely, suppose that $\epsilon_n \downarrow 0$, and for some $x$, there are $y_n \in B_{\epsilon_n}(x) \cap A$ and $z_n \in B_{\epsilon_n}(x) \cap A^c$. Then by construction $x \in \mathrm{cl}~A$ and $x \in \mathrm{cl}~A^c$ (in fact $y_n \to x$ and $z_n \to x$). Consequently,
$$
\partial A \subset \mathrm{cl}~A ~\cap\mathrm{cl}~A^c.
$$
A: Yes, $$\partial A = \overline{A} \cap \overline{A^\complement}\tag{0}$$ which is quite clear from the corrected definition (with quantifier)
$$\partial A = \{x \in M\mid \forall r>0: (B_r(x) \cap A \neq \emptyset) \land (B_r(x) \cap A^\complement \neq \emptyset)\,\}\tag{1}$$
using $$\overline{B} = \{x \in M \mid \forall r>0: B_r(x) \cap B \neq \emptyset\}$$
which holds too for all $B \subseteq M$. The $\forall$ quantifier commutes over $\land$.
Also, another way of finding the boundary is seeing it as:
$$\partial A = \overline{A} \setminus \operatorname{int}(A)$$ which is clear from
the previous equation $(0)$ plus the standard fact
$$\operatorname{int}(A)^\complement = \overline{A^\complement}$$
