Is this proof for a necessary condition for an element to belong to the boundary correct? Hi I am new to constructing a mathematical proof in topology and want to know if this argument is sound. I am a self-learner and appreciate all help. I will just do the positive statement here.
John Lee Introduction to topological manifolds.
2.8 (c)
A point is in $\partial A$ iff every neighborhood of it contains both a point of $A$ and a point of $X \setminus A$.
Assume $x \in \partial A$ and that $U$ is a neighbourhood of $x$ ($x \in U$).
$\Rightarrow x \in X\setminus \{Int(A) \cup Ext(A)\}$
$\Rightarrow \underbrace{x \notin Int(A)}_1$ and $\underbrace{x \notin Ext(A)}_2$
$1\Rightarrow x \in X \setminus Int(A)$
and since initially defined $x\in U$, $\Rightarrow X\setminus Int(A) \subseteq U$
and since $X \setminus A \subseteq X\setminus Int(A)$
(which is from the definition of interior
$Int(A) = \bigcup \{C\subseteq X : C \subseteq A$ and $C$ is open$\}$
then noting from the defn of interior, $Int(A) \subseteq A$)
can say that $X \setminus A \subseteq U$ so the neighbourhood must contain all points of $X\setminus A$ assuming it's not empty. 
I still have to do it for a point in the set $A$ and then show the converse statement is true.
My question is basically does this make sense so far?
 A: It seems (I don't have Lee's book) that he defines the $\partial A$ as $X \setminus (\operatorname{Int}(A) \cup \operatorname{Ext}(A))$, where $\operatorname{Int}(A)$ is defined as $\bigcup\{O \subseteq X: O \subseteq A \text{ and } O \text{ is open}\}$ and presumably, $\operatorname{Ext}(A) = \operatorname{Int}(X \setminus A)$.
Firstly, from $x \in U$ and $x \in X \setminus \operatorname{Int}(A)$ we don't conclude that $X \setminus \operatorname{int}(A) \subset U$. But we do know, as $U$ is open and contains $x$, we cannot have that $U \subseteq A$, because that would imply that $U$ is one of the open sets comprising $\operatorname{Int}(A)$ (see the definition) and this would mean in particular that $x \in \operatorname{Int}(A)$, which we know not to be the case.
So we know $U \nsubseteq A$, which means exactly that $U$ contains a point of $X \setminus A$. 
Now similarly, we cannot have that $U \subseteq X \setminus A$, as this would imply $x \in \operatorname{Ext}(A)$. And this exactly means that $U$ intersects $A$.
