The line in my writing symbolizes the closure, where we add all the boundary points.
Is it true that:
$\overline{A \cap B}=\overline{A}\cap {B}$?
If B is closed, but we do not know if or if not A is?(There is a simple example on the real line that shows we also must assume that A and B are not disjoint for this to possible hold.
I am only able to prove that:
$\overline{A \cap B} \subseteq \overline{A}\cap {B}$ , does it hold to other way?
Here is my proof:
$\subseteq$:
$x \in \overline{A \cap B}$
If is in $A \cap B$ we are done.
If not x is a boundary point of $A \cap B$. Note that x must be in B, because if not, since $B^c$ is open, we can find an open ball around it, and then this ball would be containd in $B^c\subseteq (A \cap B)^c=A^c \cup B^c$. And hence not be a boundary point of $A \cap B$. This only leaves that $x \in A^c$. But since every ball around x contains elements in $A \cap B$ we see that x is a boundary point of A, and hence in $\overline{A}$ and we are done.
How do I prove the other way?
You've said it yourself. If $A = [0, 1)$ and $B = [1, 2]$ then $A \cap B = \emptyset$ and $\bar{A} \cap B = {1}$. These sets needn't be disjoint. We could make $B = \{0\} \cup [1, 2]$. We can have both sets connected in $\mathbb{R}^2$, if you like.