Let A ⊂ (X,T). Then Ā = A iff A is closed. Does this it mean if we take the closure of a closed set, the result will be equal to an open set? 
I’m assuming that all adherent points of the closed set Ā exclude the boundary as the points x that lie on the boundary don’t have neighborhoods (not sure) therefore the only points will be the interior of A “the open set A” if we take the set of all adherent points of Ā.
I have stated my assumption as it might give some hints for where I am wrong. Thanks.
 A: Closure of $A$ is the smallest CLOSED set contains $A$; then $Cl(A)$ is closed by definition. On the other hand, if $A$ is closed then the smallest closed set that $A$ is subset of that set, must be $A$ itself thus $Cl(A)=A$ if and only if $A$ is closed.  
A: So the question is basically asking this:
If $(X, \mathcal{T})$ is a topological space, and $A \subseteq X$ (i.e., $A$ is a subset of $X$), then you need to prove $A$ is closed iff $\overline{A} = A$.
It seems from your notation that you think $A \subseteq (X, \mathcal{T})$ means $A$ is open.  This isn't true.  If you write $A \in \mathcal{T}$, then that means $A$ is open.  But writing $A \subseteq (X, \mathcal{T})$ just means $A \subseteq X$.
You can use multiple equivalent definitions of closure here.  
One definition of the closure of a set $A$ is $\overline{A} = A \cup A'$, where $A'$ is the set of limit points of $A$.  
You could also use $\overline{A} = \{ x \in X \mid U \cap A \neq \emptyset \text{ for all } U \in \mathcal{T} \text{ with } x \in U \}$.  That is, $\overline{A}$ is the set of $x \in X$ such that all open neighborhoods around $x$ intersect $A$.
A: I'll suppose that $\overline{A}$ is defined as the set of all adherence points of $A$ (i.e. all $x \in X$ such that for every open set $O$ that contains $x$ we have $O \cap A \neq \emptyset$), and a set is closed iff it is the complement of an open set.
So suppose $A$ is closed. We always know that all points of $A$ are adherence points of $A$ trivially (i.e. $A \subseteq \overline{A}$), so suppose that $x \notin A$. Then as $A$ is closed, $O = X \setminus A$ is open and contains $x$, but this $O$ misses $A$. This shows that $x$ is not an adherence point of $A$, or $x \notin \overline{A}$. So $x \in A$ implies $x \in \overline{A}$ and $x \notin A$ implies $x \notin \overline{A}$. So $A = \overline{A}$ as required.
Suppose then that $A = \overline{A}$. Then $A$ is closed: let $x \in X \setminus A$. Then $x \notin \overline{A}$, so $x$ is not adherence point of $A$, so there is some open set $O_x$ such that $x \in O_x$ and $O_x \cap A = \emptyset$. But this means that $O_x \subseteq X \setminus A$, and as all points of $X \setminus A$ are contained in such an $O_x$, $X \setminus A = \cup \{O_x: x \in X \setminus A\}$, which is a union of open sets, so $X \setminus A$ is open, and $A$ is closed.
A: It's almost by definition. Anyway...
$$\bar A=\bigcap_{\{C\text{ close }\mid C\supseteq A\}}C$$
but $A$ is close, and is the smallest element of $\{C\text{ close }\mid C\supseteq A\}$ therefore
$$\bigcap_{\{C\text{ close }\mid C\supseteq A\}}C=A\implies \bar A=A$$
