TITLE: Making compact subsets "parallel"
QUESTION [9 upvotes]: Let $X$ be a compact metric space. Say that two compact subsets $E,F\subset X$ are parallel if
$$ dist(x,F) = dist(y,E)$$
for all $x\in E$ and $y\in F$. Here $ dist(y,E) = \inf\{d(y,z):z\in E\}.$
The overall question I would like to understand is the following:

Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E_t\}_{t\in T}$. When is there a metric on $X$ that makes every pair of sets in $\{E_t\}$ parallel?

I generally think of $T$ as uncountable and $\{E_t\}$ as a partition of $X$, but I don't see the need to assume this.
There is at least one clear necessary condition that $\{E_t\}$ must satisfy: Namely, if $\{x_n\in E_{t_n}\}$ is a sequence converging to $x\in E_t$, then every element $x'\in E_t$ is the limit of a sequence $\{x'_n\in E_{t_n}\}$, and every sequence $\{x''_n\in E_{t_n}\}$ accumulates only at points in $E_t$.
Call this condition (*). (The part after the and in condition (*) was added after MTyson's example.)
For example, if $X=[0,1]$ and $\{E_t\}$ consists of $[0,1/2]$ as well as all the singletons $\{x\}$ for each $x\in (1/2,1]$, the collection fails condition (*) and can't be made parallel.
So I refine the question:

Let $X$ be a compact (Hausdorff, second countable, hence metrizable) topological space with a collection of disjoint compact subsets $\{E_t\}_{t\in T}$, satisfying (*). Is there a metric on $X$ that makes every pair of sets in $\{E_t\}$ parallel?

As I don't know if I believe this, I also ask:

Are there conditions, in addition to or instead of (*), that ensure the sets $\{E_t\}$ can be made parallel by the appropriate choice of a metric on $X$?

I'd be interested in counterexamples, other possible obstructions, references to similar things in the literature, or general sufficient conditions (even if they assume more about $X$ or $\{E_t\}$).
One could imagine cases in which $T$ is a topological space and $t\mapsto E_t$ is continuous in some topology on the space of compacta of $X$ (e.g., that induced by the Hausdorff metric coming from some metric on $X$.) Maybe this is equivalent to my formulation. Regardless, if that makes the problem easier or more well-known then by all means assume it.

REPLY [5 votes]: The answer to this problem is affirmative (at least for covers).
Definition. A family $\mathcal C$ of subsets of a topological space $X$ is called
$\bullet$ lower semicontinuous if for any open set $U\subset X$ its $\mathcal C$-star $St(U;\mathcal C):=\bigcup\{C\in\mathcal C:C\cap U\ne\emptyset\}$ is open in $X$;
$\bullet$ upper semicontinuous if for closed set $F\subset X$ its $\mathcal C$-star $St(F;\mathcal C)$ is closed in $X$;
$\bullet$ continuous if $\mathcal C$ is both lower and upper semicontinuous.
Theorem. For a cover $\mathcal C$ of a metrizable space $X$ by compact subsets the following conditions are equivalent:

The topology of $X$ is generated by a metric $d$ making all sets in the cover $\mathcal C$ parallel.
The cover $\mathcal C$ is disjoint and continuous.
The cover $\mathcal C$ has property $(*)$.

Proof. The implications $(1)\Rightarrow(2)\Leftrightarrow(3)$ are simple and are left to the reader. 
To prove that $(2)\Rightarrow(1)$, assume  $\mathcal C$ is disjoint and continuous. Fix any admissible metric $\rho\le 1$ on $X$.
Let $\mathcal U_0(C)=\{X\}$ for every $C\in\mathcal C$.
Claim. For every $n\in\mathbb N$ and every $C\in\mathcal C$ there exists a finite cover $\mathcal U_n(C)$ of $C$ by open subsets of $X$ such that
(i) each set $U\in\mathcal U_n(C)$ has $\rho$-diameter $\le\frac1{2^n}$;
(ii) if a set $A\in\mathcal C$ meets some set $U\in\mathcal U_n(C)$, then $A\subset\bigcup\mathcal U_n(C)$ and $A$ meets each set $U'\in\mathcal U_n(C)$.
Proof of Claim. Using the paracompactness of the metrizable space $X$, choose an open locally finite cover $\mathcal V$ of $X$ consisting of sets of $\rho$-diameter $<\frac1{2^n}$.
For every compact set $C\in\mathcal C$ consider the finite subfamily $\mathcal V(C):=\{V\in\mathcal V:V\cap C\ne \emptyset\}$ of the locally finite cover $\mathcal V$. Since the cover $\mathcal C$ is upper semicontinuous, the set $F_C=St(X\setminus \bigcup\mathcal V(C);\mathcal C)$ is closed and disjoint with the set $C$. Since $\mathcal C$ is lower semi-continuous, for any open set $V\in\mathcal V(C)$ the set $St(V;\mathcal C)$ is open and hence  $W(C):=\bigcap_{V\in\mathcal V(C)}St(V;\mathcal C)\setminus F_C$ is an open neighborhood of $C$.
It can be shown that the family $\mathcal U_n(C):=\{W(C)\cap V:V\in\mathcal V(C)\}$ satisfies the conditions (i), (ii).
Continuation of the proof of Theorem. 
Given two points $x,y\in X$ let $$\delta(x,y):=\inf\big\{\tfrac1{2^n}:\mbox{$\exists C\in\mathcal C$ and $U\in\mathcal U_n(C)$ such that $x,y\in U$}\big\}.$$
Adjust the function $\delta$ to a pseudometric $d$ letting $$d(x,y)=\inf\sum_{i=1}^m\delta(x_{i-1},x_i)$$where the infimum is taken over all sequences $x=x_0,\dots,x_m=y$.
It can be shown that the metric $d$ generates the topology of $X$ and $\rho\le d$.
Let us prove any sets $A,B\in\mathcal C$ are parallel in the metric $d$. We need to show that $d(a,B)=d(A,B)=d(A,b)$ for any $a\in A$, $b\in B$. Assuming that this equality is not true, we conclude that either $d(a,B)>d(A,B)>0$ or $d(A,b)>d(A,B)>0$ for some $a\in A$ and $b\in B$.
First assume that $d(a,B)>d(A,B)$ for some $a\in A$. Choose points $a'\in A$, $b'\in B'$ such that $d(a',b')=d(A,B)<d(a,B)$. By the definition of the distance $d(a',b')<d(a,B)$, there exists a chain $a'=x'_0,x'_1,\dots,x'_m=b'$ such that $\sum_{i=1}^m\delta(x'_{i-1},x'_i)<d(a,B)$. We can assume that the points $x'_0,\dots,x'_m$ are pairwise distinct, so for every $i\le m$ there exist $n_i\ge 0$ such that $\delta(x'_{i-1},x'_i)=\frac1{2^{n_i}}$ and hence  $x'_{i-1},x'_i\in U_i'$ for some $C_i\in\mathcal C$ and $U_i'\in\mathcal U_{n_i}(C_i)$. For every $i\le m$ let $A_i\in\mathcal C$ be the unique set with $x_i'\in A_i$. Then $A_0=A$ and $A_m=B$.
Using the condition (ii), we can inductively construct a sequence of points $a=x_0,x_1,\dots,x_m\in B$ such that for every positive $i\le m$ the point $x_i$ belongs to $A_i$ and the points $x_{i-1},x_i$ belong to some set $U_i\in\mathcal U_{n_i}(C_i)$. The chain $a=x_0,x_1,\dots,x_m\in A_m=B$ witnesses that $$d(a,B)\le  d(a,x_m)\le\sum_{i=1}^m\delta(x_{i-1},x_i)\le\sum_{i=1}^m\tfrac1{2^{n_i}}=
\sum_{i=1}^m\delta(x'_{i-1},x_i')<d(a,B),$$
which is a desired contradiction.
By analogy we can prove that the case $d(A,B)<d(A,b)$ leads to a contradiction.