Closed and bounded disjoint sets have positive distance It is well known that disjoint closed sets in $\mathbb{R}^n$ can have distance equal to 0 (take, for example, the curves $y=\pm\frac{1}{x}$). If we add the requirement that the sets be bounded, this is no longer possible, since such sets are compact by the Heine-Borel theorem.
My question is whether there exists an elementary proof of this fact. I realize that I could surreptitiously inline the relevant parts of the proof of Heine-Borel, but that's not very interesting.
While thinking about this, I came to believe that it should be slightly easier to prove this for convex sets. The key here seems to be the proposition that two closed bounded convex sets can be separated by open balls. I'm not sure how much of a simplification this is as I don't have much background in geometry; the only relevant result that comes to mind is the separation of convex sets in Banach spaces, and this looks to be way overkill. I'd appreciate any input on this as well.
 A: Let ${\rm dist}\,(A,B) = \inf \{d(x,y): x \in A, y \in B\}$ denote the distance between sets.
Suppose $A_0$ and $B_0$ are closed bounded nonempty sets in ${\mathbb R}^n$ with $d(A_0,B_0) = 0$.  $A_0$ is the union of finitely many closed bounded nonempty sets of diameter at most $1$ (e.g. take the intersections with rectangles of diameter at most $1$ that tile a large
rectangle enclosing $A_0$), and at least one of these (call it $A_1$) must have 
${\rm dist}\,(A_1, B_0) = 0$.  Similarly find $B_1 \subset B_0$ of diameter at most 1 such that ${\rm dist}\,(A_1, B_1) = 0$. 
Repeating this idea, inductively define sequences
of closed, bounded nonempty sets $A_n$ and $B_n$ with $A_{n+1} \subset A_n$, $B_{n+1} \subset B_n$, such that ${\rm dist}\,(A_n, B_n) = 0$ and $A_n$ and $B_n$ have diameter at most $1/n$.  If $x_n \in A_n$ and $y_n \in B_n$, then $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences and converge respectively to $x \in A_0$ and $y \in B_0$, and it's easy to see that $d(x,y) = 0$, so $x = y$.  Thus $A_0$ and $B_0$ are not disjoint.
