Closure of intersection is a subset of of intersection of closure - why does this method for the other way not work? I've noticed my method for proofs involving limit points isn't being used in the proofs I've read here. I'm curious if there's something I'm missing that makes them incorrect. Here's my proof for $\overline{A\cap B}\subset \bar{A}\cap\bar{B}$.
The problem reduces to showing $(A\cap B)'\subset A'\cap B'$, so let's show this:
Let $x\in(A\cap B)'\Rightarrow \forall B_r(x), B_r(x)\cap (A\cap B)\backslash\{x\}\neq \emptyset$
$\Rightarrow (B_r(x)\cap A)\cap (B_r(x)\cap B)\backslash\{x\}\neq\emptyset$
$\Rightarrow x\in A'$ and $x\in B' \Rightarrow x\in A'\cap B'$.
I know equality doesn't hold in general, but we can show the opposite since $\cap$ distributes:
Let $x\in A'\cap B'$. Then $\forall B_r(x), B_r(x)\cap A\backslash\{x\}\neq\emptyset$ and $\forall B_s(x), B_s(x)\cap B\backslash\{x\}\neq\emptyset$. 
Just take $s=r$, then we have $(B_r(x)\cap A)\cap (B_r(x)\cap B)\backslash\{x\}\neq\emptyset$.
$\Rightarrow (B_r(x)\cap A)\cap (B_r(x)\cap B)\backslash\{x\}\neq\emptyset$
$\Rightarrow B_r(x)\cap(A\cap B)\backslash\{x\}\neq\emptyset$
$\Rightarrow x\in (A\cap B)'$.
Where am I going wrong?
 A: Most easy way: ${A\cap B}\subseteq {A}\Rightarrow \overline{A\cap B}\subseteq \bar{A}$
likewise
${A\cap B}\subseteq {A}\Rightarrow \overline{A\cap B}\subseteq \bar{B}$
so
$$\overline{A\cap B}\subseteq \bar{A}\cap\bar{B}.$$
$\bar{A}\cap\bar{B}\subseteq \overline{A\cap B}$ may not happen. for example $A=Q$ and $B=Q^c$
A: You are wrong here

Let $x\in A'\cap B'$. Then $\forall B_r(x), B_r(x)\cap A\backslash\{x\}\neq\emptyset$ and $\forall B_s(x), B_s(x)\cap B\backslash\{x\}\neq\emptyset$.
Just take $s=r$, then we have $(B_r(x)\cap A)\cap (B_r(x)\cap B)\backslash\{x\}\neq\emptyset$.

The fact that
$B_r(x)\cap A\backslash\{x\}\neq\emptyset$ for all $r>0$ and
$B_s(x)\cap B\backslash\{x\}\neq\emptyset$ for all $s>0$
together doesn't implies that $(B_r(x)\cap A)\cap (B_r(x)\cap B)\backslash\{x\}\neq\emptyset$ even if you let $r=s$ since the intersection of $B_r(x)$ and $A$ might be entirely disjoint from the intersection of $B_s(x)$ and $B$
For example, let $A=(-1,0)$ and $B=(0,1)$. We can see that $0$ is a limit point of both $A$ and $B$ but any ball centered at $0$ intersects DISJOINT subset of $A$ and $B$.
