Proof that $\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}$ Given a topological space ($X$, $\mathcal{T}$) and the following defintions


*

*$\partial A$ := { $x$ | every neighbourhood of $x$ contains points from $A$ and from $X \setminus A$ }

*$\overline{A}$ := $A \cup \partial A$

*$A^{\circ}$ := $A \setminus \partial A$.


Now i want to prove that
$$
 \overline{A \cap B} \subseteq \overline{A} \cap \overline{B}
$$
Proof: With set theory
$$
 \overline{A} \cap \overline{B} = (A \cup \partial A) \cap (B \cup \partial B) 
= (A \cap B) \cup (\partial A \cap B) \cup (A \cap \partial B) \cup (\partial A \cap \partial B)
$$
and $\overline{A \cap B} = (A \cap B) \cup \partial (A \cap B)$. So if $x \in \overline{A \cap B}$. Now i distinguish two cases.
(i) $x \in (A \cap B)$, then its obvious that $x \in \overline{A} \cap \overline{B}$ with the equations from above
(ii) $x \in \partial (A \cap B)$, here i have no idea how to proceed, because i have no idea how to decompose the expression $\partial (A \cap B)$, do you have any suggestions for me?
 A: Remember that if $X\subseteq Y$ then $\overline{X}\subseteq \overline{Y}$.
As $A\cap B\subseteq A$ and  $A\cap B\subseteq B$ then  $\overline{A\cap B}\subseteq \overline{A}$ and  $\overline{A\cap B}\subseteq \overline{B}$  so
$$\overline{A\cap B}=(\overline{A\cap B})\cap(\overline{A\cap B})\subseteq \overline{A}\cap \overline{B}$$
A: You can use the following characterization of the boundary $\partial A$ of a set $A$ in a topological space $X$:
A point $x \in X$ is in $\partial A$ if and only if for every neighborhood $U$ of $x$ we have $U \cap A \neq \emptyset$ and $U \cap A^c \neq \emptyset$.
There is also an analogous definition for the closure:
A point $x\in X$ is in the closure $\overline A$ of $A$ if and only if every neighborhood $U$ of $x$ satisfies $U \cap A \neq \emptyset$.
Now apply this to $A \cap B$.
A: Let $x \in \partial (A \cap B)$. 
Let $U$ be any neighborhood of $x$. Then there exists some $y \in U \cap (A \cap B)$ and some $z \in U \backslash (A \cap B)$.
Then $y$ in $U \cap A$ and in $U \cap B$, and $z$ is either in $U \backslash A$ or in $U \backslash B$. 
Now, the problem to face is that $z$ could be in $U \backslash A$ but when you change $U$ to $U'$ you get a $z'$ in $U \backslash B$. Anyhow, looking at $U \cap U'$ should help you fix this issue....
A: The answer to the case you ask about is hidden in the definition of $\partial A$ (which is included in your question). Since it is formulated by means of neighbourhoods, you are rather supposed to use them. 
Let $\mathscr{N}_x$ be the set of all neighbourhoods of $x$.
Notice, that the following (standard) characterization of $\overline{A}$ is a consequence of your definitions:
\begin{equation}\tag{$\star$}
x\in\overline{A}\quad\text{iff}\quad(\forall U\in\mathscr{N}_x)(N\cap A\neq\emptyset).
\end{equation}
And now use ($\star$) to prove the remaining part. 
