Prove that If B is open, then $\overline{A} \cap B \subset \overline{A \cap B}$ Let $(X,d)$ be a metric space and let $A, B \subset X$. Prove that:
If B is open, then $$\overline{A} \cap B \subset \overline{A \cap B}$$ where $\overline{S}$ indicates closure for some set $S$.
For proof I was only able to change the prove part slightly(not sure if this is a correct start):
Since $B \subset \overline{B} \implies \overline{A} \cap \overline{B} \subset \overline{A \cap B}$
 A: Given a set $E \subset X$ you have $x \in \overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x \in \overline A \cap B$. Then $x \in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x \in B$ and $B$ is open, $U \cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U \cap B$ contains a point of $A$. If you denote this point by $z$, you find $z \in U \cap (A \cap B)$. Thus $U$ contains a point of $A \cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x \in \overline{A \cap B}$.
Consequently $\overline A \cap B \subset \overline{A \cap B}$.
A: Let $x\in {\overline A}\cap B$. Since $B$ is open, we can chose $r > 0$ such that $B_{r}(x)\subset B$. For some $0 < s < r$ we still have $B_s(x)\cap B\neq\emptyset$ and since $x\in\overline A$, then $B_s(x)\cap A\neq\emptyset$. Thus $B_s(x)\cap(A\cap B)\neq\emptyset$. 
We have just shown $x\in \overline{A \cap B}.$
A: The closure of a set $S$ in a metric space $X$ is the set of all points $x$ in $X$ such that there exists a sequence $(s_{n})$ in $S$ such that $\lim s_{n}=x.$ I will assume you are comfortable with this, and use it to prove the claim.
Suppose $x\in \overline{A}\cap B.$ This tells us two things: first, there exists a sequence $(a_{n})$ in $A$ such that $\lim a_{n}=x;$ and second, $x\in B.$
Since $\lim a_{n}=x$ and $B$ is open, it must be the case that for $n$ large enough, $a_{n}\in B$ also. That is, there must exist $N\in\mathbb{N}$ such that $a_{n}\in B$ for all $n\geq N.$ Define a new sequence $(c_{n})$ by the formula $c_{n}=a_{N+n}.$
For all $n\in\mathbb{N},$ we know that $c_{n}\in A,$ by choice of $(a_{n}),$ and we know that $c_{n}\in B,$ by choice of $N.$ Hence $(c_{n})$ is a sequence in $A\cap B;$ and since $(c_{n})$ is a subsequence of $(a_{n}),$ we know that $\lim c_{n}=\lim a_{n}=x.$
Therefore there exists a sequence $(c_{n})$ in $A\cap B$ such that $\lim c_{n}=x,$ meaning $x\in\overline{A\cap B},$ which is what we set out to prove.
