Prove that $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets whose union is $X$. I have been trying to prove this problem, but I am worried that my assumptions may not all be correct.
I would really appreciate it if you could offer some insight or possibly point out any assumptions  that I made that are not correct.
Notations:
$\operatorname{int}(A)$ is used for the set of all interior points of $A$, and $\partial(A)$ are the boundary points of $A$
Question
Let $A$ be a subset of a metric space $X$.
Prove that: $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets whose union is $X$.
My attempted proof
$A \cap X\backslash A = \emptyset$
Thus as $\operatorname{int}(A) \subseteq A$ and $\operatorname{int}(X\backslash A) \subseteq X\backslash A$, we have $\operatorname{int}(A) \cap \operatorname{int}(X\backslash A)= \emptyset$
This shows us $\operatorname{int}(A)$ and $\operatorname{int}(X\backslash A)$ are disjointed.
Now $\partial(A)= \overline{A} \cap \overline{X\backslash A}$
i)
$\operatorname{int}(A)\cap \partial(A)= \operatorname{int}(A)\cap\overline{A} \cap \overline{X\backslash A} = \operatorname{int}(A) \cap \overline{X\backslash A}$
Let $x\in \overline{X\backslash A}$, thus every nbd$(x)$ contains a point of $X\backslash A$
Suppose $x \in \operatorname{int}(A)$, but this will mean for any open ball $N$ of $x$,  $\ N \cap X\backslash A \neq \emptyset$
Thus $N \nsubseteq A$ making $x \notin \operatorname{int}(A)$ but this is a contradiction thus $\operatorname{int}(A)\cap \partial(A) = \emptyset$
ii)
$int(X\backslash A)\cap bd(A)= int(X\backslash A)\cap\overline{A} \cap \overline{X\backslash A} = int(X\backslash A) \cap \overline{A}$
Let $a \in \overline{A}$ thus every nbd$(a)$ contains a point of $A$
Suppose $a \in \operatorname{int}(X\backslash A)$, but this will mean for any open ball $N$ of $a$,  $\ N \cap A \neq \emptyset$
Thus $N \nsubseteq X\backslash A$ making $a \notin \operatorname{int}(X\backslash A)$ but this is a contradiction thus $\operatorname{int}(X\backslash A)\cap \partial(A) = \emptyset$
This shows that $(\operatorname{int}(A)\cup \operatorname{int}(X\backslash A))\cap \partial(A) = \emptyset$
Thus $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets
This concludes my attempted proof
 A: "Let $x\in \overline{X\backslash A}$, thus every nbd$(x)$ contains a point of $X\backslash A$
Suppose $x \in int(A)$, but this will mean for any open ball $N$ of $x$,  $\ N \cap X\backslash A \neq \emptyset$"
It's =, not $\neq$ here: $\ N \cap X\backslash A$ = $\emptyset.$
Which would be a contradiction.
And the same holds for  your second mentioned case ii)
A: Nice attempt. I would first recommend you to prove $\overline{(X\setminus A)}= X\setminus \operatorname{int}(A)$ and $\operatorname{int}(X\setminus B)=X\setminus \overline{B}$ result.
By definition of interior, $\operatorname{int}(A)\subseteq A$ and $\operatorname{int}(X-A)\subseteq X-A$. Since $X \cap (X-A)=\emptyset$, we have $\operatorname{int}(A) \cap \operatorname{int}(X-A)=\emptyset$. Here I have proved $\operatorname{int}(A) \cap \partial A=\emptyset$. Note some book define boundary as $\partial A =\overline{A}\setminus \operatorname{int}(A)$. So in that sense $\operatorname{int}(A) \cap  \partial (A) =\emptyset$ trivial. As you have shown $\operatorname{int} (X\setminus A) \cap \partial A= \operatorname{int}(X\setminus A) \cap \overline{A}$. By above result, $\operatorname{int}(X\setminus A) \cap \overline{A} =(X\setminus \overline{A})\cap \overline{A}=\emptyset$. Thus,  $\operatorname{int} (X\setminus A) \cap \partial A =\emptyset$. This completes the proof of pairwise disjoint sets.
Claim: $X= \operatorname{int}(A) \sqcup \operatorname{int}(X\setminus A) \sqcup \partial (A)$. Proof: $\operatorname{int}(A) \cup \operatorname{int}(X\setminus A) \cup \partial (A)=\operatorname{int}(A) \cup \operatorname{int}(X\setminus A) \cup (\overline{A} \cap \overline{X-A})$. By distributive law, $[\operatorname{int}(A) \cup \operatorname{int}(X\setminus A) \cup \overline{A} ] \cap [\operatorname{int}(A) \cup \operatorname{int}(X\setminus A) \cup \overline{X-A}]$. Since $ \operatorname{int}(D)\subseteq \overline{D}$, we have $[\overline{A} \cup \operatorname{int}(X\setminus A)] \cap [\overline{X-A} \cup \operatorname{int}(A)]$. Again we use above result, $\operatorname{int}(X\setminus A) =X\setminus \overline{A}$ and $\overline{X-A}=X\setminus \operatorname{int}(A)$. So $[\overline{A} \cup (X\setminus \overline{A})] \cap [(X\setminus \operatorname{int}(A)) \cup \operatorname{int}(A)] =X\cap X=X$. Hence $X =\operatorname{int}(A) \sqcup \operatorname{int}(X\setminus A) \sqcup \partial (A)$.
A: Your work is correct in showing the 3 sets are pairwise disjoint. Note this applies to any topological space $X$. Just replace the 2 occurrences of "any open ball $N$ of $x$" with "any open $N$ with $a\in N$".
If $p\in X$ but $p\not\in int(A)$ and $p\not\in int(X\setminus A)$:
(i). Every open  $V$ with $p\in V$ must satisfy $V\cap A\ne\emptyset.$ Otherwise, there exists an open $V$ with $p\in V\subset X\setminus A,$ implying $p\in V=int(V)\subset int(X\setminus A).$
So $p\in \overline A.$
(ii). Interchanging $A$ with $X\setminus A$ in (i), we also find $p\in \overline {X\setminus A}.$
(iii) Therefore  by (i) and (ii) we have $p\in\partial A.$
Other notations: $int(A)=A^o$ and $\partial A=Fr(A).$ (Fr for Frontier.)
Remark: In any topological space $X$, if $B$ is open and if $B$ is disjoint from $C$ then $B$ is disjoint from $\overline C$. Because $X\setminus B$ is closed, so $\overline C\subset \overline {X\setminus B}=X\setminus B.$
