Prove that $\partial A$ is a cutset of connected $X$ if $\operatorname{Int}(A)$ and $\operatorname{Int}(X - A)$ are nonempty Exercise 6.23 (p.202) of Introduction to Topology: Pure and Applied by C Adams and R Franzosa asks: 

Let $X$ be a connected topological space and $A$ be a subset of $X$. Prove that if $\operatorname{Int}(A)$ and $\operatorname{Int}(X - A)$ are nonempty, then $\partial A$ is a cutset, and the pair of sets, $\operatorname{Int}(A)$ and $\operatorname{Int}(X - A)$, is a separation of $X - \partial A$.

According to the book, a subset $S$ of a topological space $X$ is called a cutset of $X$ if $X - S$ is disconnected. A cutset of $X$ is said to separate $X$. 

Here is my incomplete answer: If the definition of the cutset doesn't require that a cutset be the minimum of all possible cutsets, so the proof is straight forward. I need to prove the following conjectures:


*

*$\operatorname{Int}(A)\cap \partial A = \emptyset$

*$\operatorname{Int}(X-A)\cap \partial A = \emptyset$

*$\operatorname{Int}(A)\cap \operatorname{Int}(X-A) = \emptyset$

*$\operatorname{Int}(A)\cup \partial A \cup \operatorname{Int}(X-A)= X$
I could prove 1 and 3, but unfortunately I can't prove 2 and 4. 
Could someone please help me with these questions: Is my procedure still correct if we require the cutset be the minimum of all possible cutsets? If so, how to prove 2 and 4 above?
EDIT - Is it correct to say "If $A\subset X \implies Cl(A)\subset X$"?
 A: Since $\operatorname{bdry}A=(\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)$, we automatically have $\operatorname{bdry}A=\operatorname{bdry}(X\setminus A)$. Thus, if you know that $(\operatorname{int}A)\cap\operatorname{bdry}A=\varnothing$, then you also know that
$$\big(\operatorname{int}(X\setminus A)\big)\cap\operatorname{bdry}A=\big(\operatorname{int}(X\setminus A)\big)\cap\operatorname{bdry}(X\setminus A)=\varnothing\;,$$
where the last step follows by applying $(1)$ to $X\setminus A$ instead of $A$.
More directly, $\operatorname{int}(X\setminus A)=X\setminus\operatorname{cl}A$, and $\operatorname{bdry}A\subseteq\operatorname{cl}A$, so $\operatorname{int}(X\setminus A)\cap\operatorname{cl}A=\varnothing$.
For $(4)$, note that 
$$\begin{align*}
(\operatorname{int}A)\cup\operatorname{int}(X\setminus A)&=(X\setminus\operatorname{cl}A)\cup\big(X\setminus\operatorname{cl}(X\setminus A)\big)\\
&=X\setminus\big((\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)\big)\\
&=X\setminus\operatorname{bdry}A\;.
\end{align*}$$
I usually take $(\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)$ as the definition of $\operatorname{bdry}A$; if you’re using one of the various equivalent definitions, your first step should probably be to prove that it’s equivalent to this one.

The definition of cutset does not require any kind of minimality.
