If $A$ is open then $\text{Int}(A)=A$ 
Let $(X,\tau)$ be a topological space and $A$ any subset of $X$. The largest open set contained in $A$ is called the interior of $A$ and is denoted $\text{Int}(A)$.
i$)$ Show that if $A$ is open in $(X,\tau)$ then $\text{Int}(A)=A$,

My attempt:
$a\in A$ there exists an open set $a\in\mathscr{U}_a$.
$\text{Int}(A)=\bigcup_{a\in A}\mathscr{U_a},\forall \mathscr{U}_a\subset A$.
Let's define $A'=A\cup\mathscr{U}_x$ such that $\mathscr{U}_x\nsubseteq A$, so $A'\supset A$, however since $\mathscr{U}_x\nsubseteq A$, then $\mathscr{U}_x\nsubseteq \text{Int}(A)$.
Question:
Is my proof right? If not Why? What are alternatives?
 A: There are several problems with this proof. It begins with “$a\in A$ there exists an open set $a\in\mathscr{U}_a$.” Well, this is not a sentence. What do you mean? That, for each $a\in A$, there is an open set $O$ such that $a\in O$? Sure: take $O=X$. Or, since $A$ is open, take $O=A$.
The second sentence (?) makes no sense either.
It always happens that $\mathring A\subset A$. On the other hand, if $A$ is open and if $a\in A$, then $A$ is an open subset of $X$ which is a subset of $A$ and to which $a$ belongs. Since $\mathring A$ is the union of all such subsets of $X$, $a\in\mathring A$. So, this proves that $A\subset\mathring A$. And, since $A\supset\mathring A$, $A=\mathring A$.
A: If your definition of the interior of a subset $A$ is

the largest open set contained in $A$

it is a tautology that if $A$ is an open set, it is its own interior.
A: 
"....The largest open set contained in $A$ is called the interior of $A$ and is denoted $\text{int}(A)$."

If $A$ is open then it is an open set that is contained in $A$. 
Secondly there is no set $B$ that satisfies both the conditionss


*

*1) $B\subseteq A$

*2) $B$ is properly larger than $A$ in the sense that $A\subsetneq B$
So we conclude that $A$ is the largest open set that is contained in $A$, i.e. $A=\text{int}(A)$.
