Let $(\mathbb{R}, d)$ be the real line with the usual Euclidean metric $$d(a,b)=|a-b|.$$ Can one construct a new metric $d'$ such that $d'$ induces the same topology on $\mathbb{R}$ as $d$ and $d'$ satisfies $$d'(A,B)=0 \iff \bar{A}\cap\bar{B}\neq \emptyset,$$ for all $A,B \subseteq X,$ where $$d'(A,B)=\inf \{d'(a,b) \mid a\in A, b \in B\}?$$ Clearly the original metric $d$ does not satisfy that condition: take $A=\mathbb{N}=\{1,2,3,...\}$ and $B=\{n-\frac{1}{n}\mid n \in \mathbb{N}\}.$
Thank you for your help.
As you want the same topology, we have $\lim_{k=1}^\infty n+\frac 1k=n$, hence for each $n\in\Bbb N$, we find $b_n$ with $n<b_n<n+\frac1n$ and $d'(b_n,n)<\frac 1n$. Then replacing your $B$ with $B':=\{\,b_n\mid n\in\Bbb N\,\}$, we have $\bar A\cap \bar B'=\emptyset$ and $d'(A,B')=0$.