I am quite confused by this statement, let me first post few definition that the paper I am reading, uses.
Definition: Neighborhood of a set $A$ of topological space $\mathcal{X}$ is arbitrary open set containing $A$ (Eventually, if $A$ is open, it can be a neighbourhood of itself). Neighborhood of point $x\in \mathcal{X}$ is the neighborhood of set $\{x\}$.
Definition: Assume topological space $\mathcal{X}$ and arbitrary set $A\subseteq \mathcal{X}$. A point $x\in \mathcal{X}$ satisfies one of the following:
\begin{align*}
\text{There exists a neighborhood $U$ of $x$ s.t. $U\subseteq A$}\tag{1}\\
\text{There exists a neighborhood $U$ of $x$ s.t. $U\subseteq (\mathcal{X}\setminus A)$}\tag{2}
\end{align*}
The set of all points (1), (2) respectively we denote $\operatorname{int}{A},\operatorname{ext}{A}$ and call interior, exterior of a set $A$ respectively.
Now for the theorem:
Theorem: $A$ is open if and oly if $A=\operatorname{int}{A}$.
The paper proposes the proof is very obvious, but I was trying to make sense of it. Basically it seems circular to me because whether a set is open or not depends on the actual choice of topology on $X$. Basically what I am trying to show is that $A$ is open if and only if for each point $x$ there is neighbourhood $U_x$ of $x$ s.t. $U_x\subseteq A$. So now:
Proof.
In any case $\operatorname{int}{A}\subseteq A$ is by definition.
$\Rightarrow:$ Assume $A$ is open. Take arbitrary point $x\in A$, choose $A$ as neighborhood of $x$, this means $x\in \operatorname{int}A$. This proves $A\subseteq \operatorname{int}A$ hence $A=\operatorname{int}A$.
$\Leftarrow:$ Let $A\subseteq \operatorname{int}A$. Let $x\in A$, by assumption there is a neighborhood of $x$ which also lies in $\operatorname{int}{A}$. Union all such neighborhoods. By definition, neighborhoods are open sets and thus $A$, as a union of open sets is also open.
$\blacksquare$
Would this be a good reasoning or am i making it too complicated?
Also one corollary of this should be that
Corollary: $\operatorname{int}A$ is the largest (with respect to inclusion) open set contained in $A$.
Proof. If $X\subseteq A$ is open, then by the same reasoning $X\subseteq \operatorname{int}A$ (because any neighborhood of $x\in X$ is also a neighborhood in $A$ thus lies in $\operatorname{int}A$). Union of all such neighborhoods (thus open sets) lying in $A$ forms $\operatorname{int}A$.
But why should union of all open sets in $A$ form precisely $\operatorname{int}A $?
Best Answer
To make the proof of the $\Rightarrow$ in the first theorem rigorous. Assume for each $a\in A$ there is a neighborhood $N_a\subseteq A$. Thus for each $a\in A$ is also in $\operatorname{int}{A}$. Set $\mathcal{N}:=\bigcup_{a\in A}N_a\subseteq A$. Now, we wish to show that $A\subseteq \mathcal{N}$. But this is true, because for each $a\in A$ there is a neighborhood $N_a\subseteq \mathcal{N}$ (by assumption), thus $A\subseteq \mathcal{N}$, thus $A=\mathcal{N}$. By definition, all of $N_a$ are open sets and $\mathcal{N}$ is union of open sets which is also by definition open, thus $A$ is open. Q.E.D.