Proof verification that for a subset $A\subseteq X$ in a metric space $\operatorname{int} A=X\setminus \operatorname{cl}(X\setminus A)$ This is an exercise from Conway's A Course in Point Set Topology. The following Proposition 1.1.13 (P1.1.13) is needed for my proof: Let $(X,d)$ be a metric space and $A\subseteq X$, then
a) $x\in \operatorname{int} A \Longleftrightarrow \exists r>0[B(x;r)\subseteq A]$
b) $x\in \operatorname{cl} A \Longleftrightarrow \forall r>0[B(x;r)\cap A \ne\varnothing]$
The Exercise 1.7 which is proving Proposition 1.1.15(c-2): Let $(X,d)$ be a metric space and $A\subseteq X$. Then $\operatorname{int} A=X\setminus \operatorname{cl}(X\setminus A)$.
Proof. Let $x\in \operatorname{int} A$, then we can choose a set $G\subseteq A$ with a $r_0>0$ such that the open ball $B(x;r_0)\subseteq G$. Also let $y\in B(x;r_0)$, then $y\in G$, $y\in A$, hence $y\notin X\setminus A$. As $y$ was arbitrary then we showed that $B(x;r_0)\cap(X\setminus A)=\varnothing$, which by P1.1.13(b) implies that $x\notin \operatorname{cl} (X\setminus A)$. Now, since $x\in \operatorname{int} A\subseteq A\subseteq X$, then $x\in X\setminus \operatorname{cl}(X\setminus A)$. Given that $x$ was arbitrary, thus $\operatorname{int} A\subseteq X\setminus \operatorname{cl}(X\setminus A)$.
Let $x\in X\setminus \operatorname{cl}(X\setminus A)$. Given $x\notin \operatorname{cl}(X\setminus A)$ by P1.1.13(b) we can choose an $r_0>0$ such that $B(x;r_0)\cap(X\setminus A)=\varnothing$, i.e., $\forall y(y\in B(x;r_0)\implies(y\in X\implies y\in A))$, thus $B(x;r_0)\subseteq A$. By P1.1.13(a) $B(x;r_0)\subseteq A$ implies that $x\in \operatorname{int} A$. Hence, as $x$ was arbitrary, $X\setminus \operatorname{cl}(X\setminus A)\subseteq \operatorname{int} A$.
$X\setminus \operatorname{cl}(X\setminus A)= \operatorname{int} A$. $\square$
Is this proof right?
Thank you.
 A: The proof looks good to me. If you wish, you may clean it up a bit to make it concise and more readable. In a typical point-set topological proof, so many details are not required. Also, I found certain parts of your proof to be unnecessary. Here's how I would do it:

Let $x\in \operatorname{int} A$. There exists $r_0>0$ such that $B(x;r_0)\subseteq A$. Thus, $B(x;r_0)\cap(X\setminus A)=\varnothing$, which by P1.1.13(b) implies that $x\notin \operatorname{cl} (X\setminus A)$. Therefore, $x\in X\setminus \operatorname{cl}(X\setminus A)$. In conclusion, $\operatorname{int} A\subseteq X\setminus \operatorname{cl}(X\setminus A)$.

Note that we don't need to introduce the set $G$, as you have done in your proof.

Let $x\in X\setminus \operatorname{cl}(X\setminus A)$. As $x\notin \operatorname{cl}(X\setminus A)$, by P1.1.13(b) there exists $r_0>0$ such that $B(x;r_0)\cap(X\setminus A)=\varnothing$. Thus, $B(x;r_0)\subseteq A$. By P1.1.13(a), $x\in \operatorname{int} A$. Hence, $X\setminus \operatorname{cl}(X\setminus A)\subseteq \operatorname{int} A$.

