Suppose $A\subset B\subset C$ and $\overline{C}=A^{\circ}$. Show $B$ is open and closed.
I'm relatively new to topology and the following is what I "THINK" I've proven but not exactly sure. I would really appreciate if someone corrected me if I'm wrong or if there is more direct way to do this as I don't feel too confident in my proof.
For reference these are the definitions I was given and therefore been using for the proof.
$x\in\overline{A}\iff\forall(G\in\tau, x\in G),(G\cap A\neq\emptyset=\exists y\in G\cap A)$.
$x\in A^\circ\iff\exists G\in\tau,x\in G\subset A$.
$x\in A^a\iff\forall(G\in\tau,x\in G),\exists y\neq x\in G\cap A$.
$\bf{Proof:}$ Assume $A\subset B\subset C$ and $\overline{C}=A^\circ$.
$\overline{C}=A^\circ\Rightarrow \overline{C}\subset A^\circ$ and $A^\circ\subset\overline{C}$.
Consider $x\in A^\circ$, then $\exists G\in\tau$ such that $x\in G\subset A$.
$x\in A\Rightarrow x\in\overline{C}\Rightarrow\exists y\neq x\in G\cap C\Rightarrow y\in G$ and $y\in A$.
Since $G\subset A\Rightarrow G\subset A\subset B\subset C$ and $y\in G$ then $y\in G\subset A\subset B\subset C$ in particular $y\in B$.
Because $y\in G$ and $y\in B\Rightarrow \exists y\neq x\in G\cap B$ then $x\in B^a$.
Then $x\in G\subset A\Rightarrow x\in A\subset B\subset C$ in particular $x\in B$ then $B^a\subset B\Rightarrow B$ is closed.
I'm not sure if I'm using the definitions correctly but so far this is what I think I've proven. If this is actually correct, I have no clue how to prove that $B$ is open now. I would appreciate any advice. Thanks.
$\bar{C} =A^{\circ} \subset A \subset B \subset C \subset \bar{C} $ shows that all inclusions are actually equalities. Then $B=\bar{C}=A^{\circ}$ which is indeed closed and open.