Making compact subsets "parallel" 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.
 A: 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.
A: If $(E_t)_{t\in T}$ is a partition of a metric space $(X,d)$ into parallel compact subsets, then the metric $d$ induces a quotient metric $\hat d$ on the index set $T$. This metric is defined by $\hat d(t,s)=\inf\{d(x,y):x\in E_t,\;y\in E_s\}$ for $t,s\in T$. 
It is easy to show that the quotient map $q:X\to T$ (defined by $q^{-1}(t)=E_t$) is open and perfect. We recall that a map $f:X\to Y$ is perfect if it is continuous, closed and has compact preimages of points.
So, in case of partitions, the original problem reduces to the following one:
Problem. Let $f:X\to Y$ be an open perfect map between metrizable spaces. Is the topology of $X$ generated by a metric making the fibers of the map $f$ parallel?
For zero-dimensional space $X$ the affirmative answer to this problem is given by the following
Theorem 0. For any open perfect map $f:X\to Y$ between metrizable spaces the space $X$ admits an ultrametric $d$ that generates the topology of $X$ and makes the fibers $f^{-1}(y)$, $y\in Y$, of the map $f$ parallel.
Proof of Theorem 0. Fix any metric $\rho$ generating the topology of the space $X$. 
Let $\mathcal U_0=\{X\}$. 
Claim. There exists a sequence $(\mathcal U_n)_{n=0}^\infty$ of disjoint open covers of $X$ such that $\mathcal U_0=\{X\}$ and for every $n\in\mathbb N$ the following conditions are satisfied:


*

*Each $U\in\mathcal U_n$ is contained in some $V\in\mathcal U_{n-1}$;

*$\max_{U\in\mathcal U_n}diam_\rho(U)<\frac1n$;

*For any points $y,z\in Y$ if $f^{-1}(z)$ intersects some set in the family $\mathcal U_n(y):=\{U\in\mathcal U_n:U\cap f^{-1}(y)\ne\emptyset\}$, then $f^{-1}(z)$ intesects each set in the family $\mathcal U_n(y)$.
Proof of Claim. The Claim is proved by induction. Assume that for some $n\in\mathbb N$ the disjoint cover $\mathcal U_{n-1}$ has been constructed. Since the space $X$ is zero-dimensional, there exists a disjoint open cover $\mathcal V$ of $X$ such that each set $V\in\mathcal V$ has $\rho$-diameter $<\frac1n$ and is contained in some set $U\in\mathcal U_{n-1}$. For every $y\in Y$ consider the family $\mathcal V_y=\{V\in\mathcal V:V\cap f^{-1}(y)\ne\emptyset\}$, which is finite by the compactness of the fiber $f^{-1}(y)$. Since the map $f$ is open and closed, there exists a clopen neighborhood $W_y\subset Y$ of $y$ such that for each $z\in W_y$ the set $f^{-1}(z)$ is contained in $\bigcup \mathcal V_y$ and intersects each set $V\in\mathcal V_y$. Such neighborhood $W_y$ can be defined as $W_y:= \bigcap_{V\in\mathcal V_y}f(V)\setminus f(X\setminus\bigcup\mathcal V_y)$. Since the space $Y$ is zero-dimensional, there exists a disjoint open cover $\mathcal W$ of $Y$, refining the cover $\{W_y\}_{y\in\mathcal Y}$. It is easy to check that the cover $\mathcal U_n=\{V\cap f^{-1}(W):V\in\mathcal V,\;W\in\mathcal W\}$ has the properties (1)--(3). 
Now let us return to the proof of Theorem 0. On the space $X$ consider the ultrametric $d$ defined by $$d(x,y)=\inf\Big\{\frac1{2^n}:\exists U\in\mathcal U_n\; x,y\in U\Big\}.$$ The condition (1)
 of Claim guarantees that $d$ is an ultrametric, the condition (2) ensures that the ultrametric $d$ generates the topology of $X$, and the condition (3) guarantees that the fibers of the map $f$ are parallel in the metric $d$.
Remark. The problem posed above has an affirmative answer, given in my next post.
A: No, (the old version of) $(*)$ alone doesn't imply that there is a metric making the sets parallel.
Take $X=[-1,1]$ and let $\{E_t\}=\{-t,t\}_{0\le t<1}\cup\{-1\}\cup\{1\}$. It's easy to see that $(*)$ is satisfied here. Suppose there were a metric making $d$ making the sets parallel. Then $d(\frac{n-1}{n},1)=d(-\frac{n-1}{n},1)$ yet $\frac{n-1}{n}\to 1$ and $-\frac{n-1}{n}\to -1$, so $d(-1,1)=0$.
To get around this counterexample, you could try assuming the leaves are connected or that $T$ (with the quotient topology?) is Hausdorff.
