= Solution
For nonempty $F$, the <triangle inequality> gives $d(x,z)\leq d(x,y)+d(y,z)$ for every $z\in F$. Taking the infimum yields $d(x,F)\leq d(x,y)+d(y,F)$. Interchanging $x,y$ proves
$$
\boxed{|d(x,F)-d(y,F)|\leq d(x,y).}
$$
Thus the <distance to a set> is a <Lipschitz function>, hence continuous. If $x\notin F$ and $F$ is closed, its open complement contains a ball $B(x,\epsilon)$; every point of $F$ is at least $\epsilon$ away. Hence $\boxed{d(x,F)>0}$ for $x\notin F$.
For two nonempty disjoint closed sets, let $h(x)=d(x,F_1)-d(x,F_2)$. It is continuous, negative on $F_1$ and positive on $F_2$. The disjoint open sets $U_1=\{h<0\}$ and $U_2=\{h>0\}$ contain them. Empty sets cause no difficulty, using an empty neighborhood for the empty set. This proves \b[every metric space is a <normal topological space>].
In any normal space, a closed $F$ contained in an open $U$ can be separated from $X\setminus U$: choose disjoint open $W\supset F$ and $V\supset X\setminus U$. Because $X\setminus V$ is closed and contains $W$, $\overline W\subset X\setminus V\subset U$. First choose disjoint open $U_1,U_2$ around $F_1,F_2$, then apply this argument separately to each pair $F_i\subset U_i$. It gives $\overline W_i\subset U_i$, and therefore $\boxed{\overline W_1\cap\overline W_2=\varnothing}$. This is <closed-neighborhood shrinking in a normal space>.
For the two specified hyperbola branches, explicit choices are
$$
\boxed{W_1=\{(x,y):x<0,\ xy<-1/2\},\qquad W_2=\{(x,y):x>0,\ xy>1/2\}.}
$$
They are open and contain the branches with products $-1$ and $1$. Their closures lie respectively in $\{x\leq0,xy\leq-1/2\}$ and $\{x\geq0,xy\geq1/2\}$. Any common point would have $x=0$, making both product inequalities impossible. Thus their closures are disjoint even though the original branches can approach one another arbitrarily closely at large height.
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