Consider the following property in a metric space $(X,d)$: If $A,B$ are closed and disjoint, then $dist(A,B)>0$. Does this imply $(X,d)$ is complete?
My guess is yes. I begin by considering a Cauchy sequence $(x_n)$ in $X$ that does not converge. Then, for every fixed $x_0$ in $X$ and every $\epsilon>0$, the ball $B(x_0,\epsilon)$ must contain zero or finite points of $(x_n)$. This gives us a set $A=cl(B(x_0,\epsilon))$, cl being its closure. How could I construct B?
I am thinking on those lines: if $(x_n)$ is a Cauchy sequence that doesn't converge, it cannot even have a subsequence that converges, because then it would be easy to conclude that the whole sequence converges (to the same limit). In particular, it can take every value only finitely many times, so WLOG we can assume all its values are mutually different. Also, all its subsequences are Cauchy sequences.
Now, what about the set $\{x_n, n\in\mathbb N \} $. I would argue that it is closed. Otherwise, a point $x $ in its closure but not in the set would be a limit point, and so the sequence would have a convergent subsequence.
With all this said, let's now just take $A=\{x_{2n}, n\in\mathbb N\} $, $B=\{x_{2n+1}, n\in\mathbb N\} $. Those are disjoint (remember we could assume $x_n $ are all different), closed and at distance $0$ - a contradiction.