Why does every neighborhood of a boundary point contain an element of the set it is bounding and the space minus the set. I am trying to prove that every neighborhood of a boundary bound contains a point in interior and $X \setminus A$ where $A$ is the set in consideration. I am given the following definitions
(1) $(X,\mathcal{T})$ is a topological space if $X,\emptyset \in \mathcal{T}$, any arbitrary union of open sets in $\mathcal{T}$ is an open set, and any finite intersection of an open set in $\mathcal{T}$ is an open set.
(2) A neighborhood of a subset of $X$ is an open set containing the subset.
(3) A set $A \subseteq X$ is closed if $X \setminus A$ is open
(4) $\bar A$ = $\cap\{ B \subseteq X : A \subseteq B \text{ and } B \text{ is closed in X} \}$
(5) Int($A$) = $\cup \{ B \subseteq A : B \text{ is open in } X \}$ 
(6) Ext($A$) = $X \setminus \bar A$
(7) $\partial A$ = $X \setminus ( \text{Int(}A)\text{)}\cup \text{Ext(}A \text{)})$
From this I can derive the closed definition of a topology and that $\partial A$ is closed. I am having trouble seeing how the previous facts and definitions give me that every neighborhood of a point in the boundary contains a point in $A$ and $X \setminus A$ where $X$ is the space and $A$ is some subset of the space. What are some derivations that would help me figure this out?

What I think it comes down to is showing that given an open neighborhood $\mathcal{O}$ of some point $x \in \partial A$, $\mathcal{O} \cap Int(A)\ne \emptyset$ and $\mathcal{O} \cap (X \setminus A) \ne \emptyset$
 A: Observe that if $x$ is in the boundary of $A$, then $x$ is not in the interior or the exterior of $A$. Since $x\notin\operatorname{int}A$, $x$ has no open nbhd contained entirely in $A$. But that just means that if $U$ is an open nbhd of $x$, then $U\nsubseteq A$, and hence $U\cap(X\setminus A)\ne\varnothing$. A similar argument shows that since $x$ is not in the exterior of $A$, every open nbhd of $x$ must meet $A$ in a non-empty set. The converse is equally easy to show: if every open nbhd of $x$ meets both $A$ and $X\setminus A$, then $x$ is in the boundary of $A$. To complete the argument, prove the following proposition:

Proposition. For any set $S\subseteq X$ and any $x\in X$, $x\in\operatorname{cl}S$ if and only if $U\cap S\ne\varnothing$ for each open nbhd $U$ of $x$.

A: Using the definitions straightforwardly:
Let $x\in\partial A$, and let $B$ be a neighbourhood of $x$. Suppose that $A\cap B=\emptyset$. This implies that $x$ is an interior point of $X\setminus A$: indeed, $A\subset X\setminus B$, which is a closed set; as $x\not\in X\setminus B$, this shows that $x\not\in\bar A$. But this would imply $x\in\mbox{Ext}(A)$ and $x\not\in\partial A$. This contradiction shows that $A\cap B\ne\emptyset$. 
If $(X\setminus \bar A)\cap B=\emptyset$, then $y\not\in \mbox{Ext}(A)$ for every $y\in B$. This implies that $B\subset A$, which makes $x$ an interior point of $A$. But this contradicts the fact that $x\in \partial A$. We conclude that $(X\setminus\bar A)\cap B\ne\emptyset$. 
