Distance between sets implies completness of a metric space 
Given the metric space $(X,d)$ suppose that for every nonempty, closed $A,B\subseteq X$ with $$A\cap B=\varnothing, \mathrm{dist}(A, B)>0.$$
  Show that  $(X,d)$ is complete.

But given the definition of distance as $$\operatorname{dist}(A,B)=\inf\{d(a,b):a \in A,b \in B\}$$ 
and given that  $$A\cap B=\varnothing$$ 
and the fact that $A,B$ are closed 
we have that  $$\mathrm{dist}(A, B)=0\iff A\cap B \neq \varnothing$$


Am I missing something?

As Micapps pointed out in the comments, in $\mathbb{Q}$ with the usual metric, $[0,\sqrt{2}), (\sqrt{2},5]$ are disjoint closed sets with distance zero, so your last statement is not true.
To show the problem, we show the contrapositive. Let $\{s_n\}_{n\in \mathbb{N}}$ be a non-convergent Cauchy sequence of distinct points (which thus has no accumulation point). For each $s_{2n}$, let $B_{2n}$ be an open ball about $s_{2n}$ that does not include any of the other $s_i$ ($i \neq 2n$). (In particular, none of these open balls include any odd-indexed $s_i$.)
Let $A = \{s_{2n} : n \in \mathbb{N}\}$ and $B = \{x : x\notin B_{2n}$ for any $n\}$, i.e. the complement of the union of the $B_{2i}$. Then $A,B$ are disjoint closed sets with distance $0$.
