 Solution

For nonempty $F$, the triangle inequality gives $d(x,z)\le d(x,y)+d(y,z)$ for every $z\in F$. Taking the infimum yields $d(x,F)\le d(x,y)+d(y,F)$. Interchanging $x,y$ provesThus 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 $B(x,ϵ)$; every point of $F$ is at least $ϵ$ 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 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 areThey are open and contain the branches with products $-1$ and $1$. Their closures lie respectively in $\{x\le 0,xy\le -1/2\}$ and $\{x\ge 0,xy\ge 1/2\}$. 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.