Let $(X,d)$ be a metric space such that $d(A,B)>0$ for any pair of disjoint closed subsets $A,B\subset X$. Show that $(X,d)$ is complete.
Suppose $X$ is not complete. Then there exists a Cauchy Sequence in $X$ which does not converge. Let $x_n$ be a Cauchy sequence of distinct terms.
How to find sets $A$ and $B$? Should I use $A=\{x_{2k}:k\in \mathbb N\}$ and $B=\{x_{2k+1}:k\in \mathbb N\}$ then $A\cap B=\phi$.
I can't proceed further.
A contradiction proof isn't necessary, but your idea is on the right track. Recall that to prove a Cauchy sequence converges, it suffices to find a convergent subsequence. Let $(x_n)$ be a Cauchy sequence. If $D:=\{x_n\,:\,n\in\mathbb N\}$ is finite then there exists $x\in X$ such that $x_n=x$ for infinitely many $n$. This implies $x$ is the limit of a subsequence of $(x_n)$. Hence assume $D$ is infinite, so we have a subsequence $(x_{n_k})$ such that $x_{n_j}=x_{n_k}$ implies $j=k$. Now your idea comes into play. Define $$A=\{x_{n_{2k}}\,:\,k\in\mathbb{N}\},\quad B=\{x_{n_{2k+1}}\,:\,k\in\mathbb N\}.$$ These are disjoint sets by choice of the subsequence $(x_{n_k})$. Since this subsequence must also be Cauchy, for every $\epsilon>0$ there exists $x\in A,y\in B$ such that $d(x,y)<\epsilon$. Hence $d(A,B)=0$, so by the given condition we cannot have that $A$ and $B$ are both closed. Assuming without loss of generality that $A$ is not closed, then take $x\in\overline A\setminus A$ and show that $x$ must be the limit of a subsequence of $(x_{n_{2k}})$. This completes the proof.
