Consider $n$ dimensional Euclidean space $\mathbb{R}^{n}$, its topology is usual topology, which induced from distance $d(x,y)=\sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}$. In topology space $(\mathbb{R}^n,d)$, a sequence $a_n$ converges to a point $a\in\mathbb{R}^n$. Given a subset $A=\{a_n:n=1,2,\ldots\}\subset\mathbb{R}^n$, its closure is $\bar{A}=A\cup\{a\}$ by definition. Another subset $B\subset\mathbb{R}^n$ does not intersect with $\bar{A}$. Does the distance $$\inf\{d(x,y):x\in A,y\in B\}>0$$ always hold?
I tried to set $A=\{(\frac{1}{n},0,\ldots,0)\}$ and $B=(\frac{1}{n}+\frac{1}{2^n},0,\ldots,0)$, although $\inf d(x,y)=\inf\sqrt{\frac{1}{2^{n-1}}}=0$, but I found if $n\to\infty$, $B$ tends to $\{0\}\subset\bar{A}$.
Edited:
By the below answer we found a counterexample when $B$ is open set. I was wondering the situation when $B$ is closed set. Two closed subsets in $\mathbb{R}^n$ do not intersect, maybe by $\mathbb{R}^n$ is Hausdorff, the distance $d(A,B)$ cannot be zero.
For simplicity, set $n=1$, then consider $A=\{\frac{1}{n}\mid n\in\mathbb{N}\}$, so that $\overline A=\{0\}\cup A$, and $B=\{1+\frac{1}{n}\mid n\in\mathbb{N}\}$, so that $\overline{A}\cap B=\emptyset$, but it is still easy to see that $d(A,B)=0$.
edited after OP's comment Another counterexample can be obtained considering $A=\mathbb{N}$, so that $\overline{A}=A=\mathbb{N}$, and $B=\{n+\frac{1}{2n}\mid n\in\mathbb{n}\}$, so that $\overline{B}=B$ and $A\cap B=\emptyset$. In this case, one can still see that $d(A,B)=0$.