Covering a compact set with balls whose centers do not belong to other balls. Let $K\subset \Bbb R^n$ be a compact set such that each $x\in K$ is associated with a positive number $r_x>0$.

Claim: $K$ can be covered by a family of balls 
  $$
\mathcal B = \{ B(x_i,r_i) : i=1,\dots,k\ \},
$$
  where $r_i := r_{x_i}$, such that for any distinct $i,j \le k$, we have 
  $$
x_i\notin B(x_j,r_j) \quad\text{and}\quad x_j\notin B(x_i,r_i).
$$

Is the claim true without any additional assumptions on $r_x$'s?
At first I thought of using Zorn's lemma to extract a maximal subfamily of balls from $\mathcal F = \{B(x,r_x/2) : x\in K \}$ such that any pair of balls is disjoint. However, enlarging the radii by a factor of $2$ may not be a cover of $K$ so this approach may not work.
 A: This is an attempt (I hope successful) at a positive answer, at least for $n=1$, i.e. when the compact set $K$ is a subset of the real line $\Bbb R^1$. It relies on the linear order of the line, and I do not immediately see how to generalize to $\Bbb R^n$, for $n\ge2$. By the way, the concept discussed in this question resembles the notion of so-called $D$-space in general topology, and in that context the Sorgenfrey line (i.e. the real line with the upper limit topology) is an important example. The class of $D$-spaces was introduced by van Douwen and Pfeffer in 1978:  
Some properties of the Sorgenfrey line and related spaces.
Washek F. Pfeffer and Eric K. van Douwen
Pacific J. Math. Volume 81, Number 2 (1979), 371-377.
https://projecteuclid.org/euclid.pjm/1102785280
and it remains an open question if every paracompact space is a $D$-space. There are variations of this concept, including strong $D$-space, and these properties are defined using so-called open neighborhood assignments, ONA, for a topological space $X$: For each point $x$, an open neighborhood $U_x$ containing $x$ is assigned. Then one picks a so-called kernel: A subset $D$ of $X$ so that the neighborhoods assigned to points in the kernel cover the space $X$. If one could always pick a kernel that is closed-and-discrete, then $X$ is called a $D$-space. (Here "always" means for every ONA, i.e. every assignment of neighborhoods $U_x$.) If a certain additional condition is satisfied (namely, the kernel $D$ is locally-finite in the topology generated by the ONA), then the space $X$ is called strongly $D$ (and this latter property appears to be fairly restrictive). In the present question, one asks for even more: each point in the kernel is contained in only its own neighborhood, and not in the neighborhoods assigned to other points in the kernel. But, at least, we work in a compact metric space, or even a compact subset of $\Bbb R^n$, and to each $x$ we assign an open ball centered at $x$, not an arbitrary neighborhood (the latter option tends to make the problem less tractable). I might have had some advantage, being familiar with some of the results on $D$-spaces (and in particular, about the Sorgenfrey line). 
I think I have a yes answer to the OP when $n=1$, i.e. when the compact $K$ is a subset of $\Bbb R^1$. I do not immediately see how to generalize to $\Bbb R^n$ for $n\ge2$ (though, for $D$-spaces, there is a paper by de Caux:
Peter de Caux, Yet another property of the Sorgenfrey plane,
Top. Proc. 6, no 1 (1981) pp. 31-43.   http://www.topo.auburn.edu/tp/reprints/v06/tp06105s.pdf
that might have relevant ideas or techniques, that deals with powers $S^n$ of the Sorgenfrey line $S$, showing that each such power is a (hereditarily) $D$-space). 
Let $K$ be a compact subset of the real line, and let for each $x\in K$ a positive radius $r_x$ is fixed, so 
each $x$ is covered by the ball $B(x,r_x)$. 
Let $x_0=\min K$ and by transfinite recursion, if 
$1\le\alpha<\omega_1$ and $x_\beta$ are defined for all $\beta<\alpha$, let $U_\alpha=\cup_{\beta<\alpha}B(x_\beta,r_{x_\beta})$ 
and let $Y_\alpha=K\setminus U_\alpha$. 
If $K\not\subseteq U_\alpha$, i.e. if $Y_\alpha\not= 
\emptyset$, then let $x_\alpha=\min Y_\alpha$. (Use that each $Y_\alpha$ is compact and has a smallest element, i.e. $\min$ with respect to the usual order of the reals.) There is a smallest countable ordinal $\gamma\ge1$ such that $Y_\gamma=\emptyset$, i.e. 
$K\subseteq U_\gamma$ but $K\not\subseteq U_\alpha$  if $\alpha<\gamma$ (where $U_0=\emptyset$). 
Claim 1. $\gamma$ is a successor ordinal, i.e. $\gamma=\delta+1$ for some $\delta<\omega_1$. 
Proof. If $\gamma$ were a limit ordinal, then $K$ would be a strictly increasing union of the open sets 
$U_\alpha$, $\alpha<\gamma$ (more precisely, 
a strictly increasing union of the relatively open sets 
$U_\alpha\cap K$, since 
$x_{\alpha+1}\in U_{\alpha+1}\setminus U_\alpha$ for each $\alpha<\gamma$). 
That is, the cover 
$\{U_\alpha:\alpha<\gamma\}$ of $K$ would have no finite subcover, a contradiction. 
Let $X=\{x_\alpha:\alpha<\gamma\}$. By Claim 1, 
$X=\{x_\alpha:\alpha\le\delta\}$. 
Note that Claim 1 and the above construction imply that $\max K\in B(x_\delta, r_{x_\delta})$. 
Claim 2. $X$ is well-ordered as a subset of $K$ (or 
equivalently, as a subset of $\Bbb R$). 
Proof. It is immediate from the construction that $X$ 
is order-isomorphic to $\gamma$, where as usual 
$\gamma=\{\alpha:\alpha<\gamma\}=[0,\gamma)$. 
If $X\subseteq B(x_\delta, r_{x_\delta})$ then we 
are done. If not then we continue as follows. 
Claim 3. If the set $P=X\setminus B(x_\delta, r_{x_\delta})$ is non-empty, then it has a maximal element. 
Proof. Note that $P$ is an initial segment of $X$, 
say $P=\{x_\alpha:\alpha<\mu\}$ for some countable ordinal $\mu\le\delta$. If $P$ did not have a maximal element, then $\mu$ is a limit ordinal. For each $\alpha<\mu$ we have $U_\alpha\subseteq(-\infty,x_\alpha)$. 
Let $q=\sup P = \sup_{\alpha<\mu} x_\alpha$. Then 
$q\in\overline P\subseteq K$ and 
$U_\mu=\cup_{\alpha<\mu}U_\alpha\subseteq(-\infty,q)$. Since
$q=\sup P=\sup U_\mu=\min (K\setminus U_\mu)$, 
we have $q=x_\mu$. 
Note that $x_\alpha<\inf B(x_\delta, r_{x_\delta})$ 
for all $\alpha<\mu$, 
hence $q=x_\mu\le\inf B(x_\delta, r_{x_\delta})$, 
and $x_\mu\not\in B(x_\delta, r_{x_\delta})$, 
implying $x_\mu\in P$, a contradiction, which completes the proof of Claim 3.  
Let $z_0=x_\delta$ and $C_0=B(z_0, r_{z_0})$. 
If $X\setminus C_0\not=\emptyset$, then (using Claim 3) let $z_1=\max (X\setminus C_0)$. 
Note that (as is easy to verify) if 
$C_1=C_0\cup B(z_1, r_{z_1})$ then 
$[z_1,\infty)\cap K\subseteq C_1$. 
If $X\not\subseteq C_1$ then let 
$z_2=\max (X\setminus C_1)$ (the proof that $\max$ exists is similar to Claim 3), and let 
$C_2=C_1\cup B(z_2, r_{z_2})$. 
Inductively, $z_{n+1}=\max (X\setminus C_n)$ 
and $C_{n+1}=C_n\cup B(z_{n+1}, r_{z_{n+1}})$, 
with $z_{n+1}<z_n$ and 
$[z_{n+1},\infty)\cap K\subseteq C_{n+1}$. 
Since the $z_n$ form a decreasing sequence 
in the well-ordered set $X$, this process 
must terminate in finitely many steps, i.e. 
there is $m\ge0$ such that $X\subseteq C_m$. 
Then also $K\subseteq C_m$, since $C_m$ covers 
everything in $K$ starting at the top $x_\delta$ 
(with $C_0=B(z_0, r_{z_0})=B(x_\delta, r_{x_\delta})$ covering $\max K$) and going back to the bottom $x_0=\min X =\min K$, without omitting 
any elements of $K$ in between. (Exercise.) 
Then the family $\mathcal B = \{ B(z_n,r_{z_n}) : n=0,\dots,m\ \}$ shows that the Claim in OP holds. 
This answers the question positively for the case when 
$K$ is a compact subspace of the real line $\Bbb R$. 
Edit. At hindsight the proof could be more instructive and unified if one first proves the following Claim 0, which could then be applied in the proofs of both Claim 1 and Claim 3 above. 
Claim 0. $X$ is closed (as a subspace of $K$ and of $\Bbb R$). 
Proof. It is more or less done in each of the proofs of Claim 1 and Claim 3, but here is 
how it could go. Since $X$ is well-ordered, 
it is enough to show that whenever $X$ contains 
an increasing sequence $a_0 < a_1 < \dots$ 
where $a_n=x_{\beta_n}$ with 
$\beta_n < \beta_{n+1}$ for all natural numbers $n<\omega$, and if we let $a=\sup_n a_n$ 
then  $a\in X$. 
Indeed, $a_n=x_{\beta_n}\in 
U_{\beta_n+1}$ 
and $U_{\beta_n}\subseteq (-\infty,x_{\beta_n})$, 
hence if $\beta=\sup_n \beta_n$ then 
$U_\beta=\cup_{n<\omega}U_{\beta_n}\subseteq 
(-\infty,a)$. Since $K$ is closed and 
$a_n\in X\subseteq K$ we have $a\in K$, and 
$a=\min (K\setminus U_\beta)$, hence 
$a=x_\beta\in X$, showing that $X$ is closed, 
completing the proof of Claim 0. (In particular, $X$ as a subspace of $K$ and of $\Bbb R$ is not only order-isomorphic to $\lambda$ but also 
homeomorphic to it.) 
With the aid of Claim 0, here are some more details justifying that the 
family $\mathcal B = \{ B(z_n,r_{z_n}) : n=0,\dots,m\ \}$ is indeed a cover of $K$. 
As above, let $z_0=x_\delta$ and 
$C_0=B(z_0, r_{z_0})$. 
By construction we have that 
$\max K\in B(x_\delta, r_{x_\delta})=C_0$. 
Since $X$ is well-ordered, we could let 
$y_0=\min(X\cap C_0)$. 
If $y_0\not=x_0=\min K$, that is if 
$X\not\subseteq C_0$ or equivalently 
$X\setminus C_0\not=\emptyset$ 
then let 
$z_1=\sup(X\setminus C_0)$. Since (by Claim 0) 
$X$ is closed, we have $z_1\in X$. But $z_1\not\in C_0$ since $C_0$ is open, hence 
$z_1<y_0$, and $y_0$ is the immediate successor 
of $z_1$ in the well-order $X$. It follows 
(by construction of $X$) that 
$y_0=\min (K\cap(z_1,\infty)\setminus 
B(z_1, r_{z_1}))$ so that there are no 
elements of $K$ "between" $B(z_1, r_{z_1})$ 
and $B(z_0, r_{z_0})$ (that is, between 
$z_1$ and $z_0$) that are not covered by 
the set 
$C_1=B(z_1, r_{z_1})\cup B(z_0, r_{z_0})=
B(z_1, r_{z_1})\cup C_0$. 
We can continue: 
Let $y_1=\min(X\cap C_1)$. 
If $y_1\not=x_0$, that is if 
$X\not\subseteq C_1$ or equivalently 
$X\setminus C_1\not=\emptyset$ 
then let 
$z_2=\sup(X\setminus C_1)$. 
Then $z_2\in X$ and $y_1$ is the immediate successor of $z_2$ in $X$, etc. 
A: I think @Fedja's counterexample on the MO counterpart of the question works pretty well. His/her answer is worthy of mentioning here:

This trivial counterexample in $\mathbb R^2$ should have taken me five minutes. Instead, I spent almost two days. The moral is the usual one: after 50 you'd better give up on mathematics.
Let $y,z$ be 2 points at distance $1$ from each other. We shall construct by induction a sequence of points $x_j$ and radii $r_j>\max(d(x_j,y),d(x_j,z))$ such that $x_j\to y$ when $j$ is odd, $x_j\to z$ when $j$ is even, $x_j$ do not lie on the line $yz$, $\max(d(x_j,y),d(x_j,z))<1$, the disk $D(x_j,\max(d(x_j,y),d(x_j,z)))$ contains  $x_1,\dots, x_j$ but $x_{j+1}\notin D(x_j,r_j)$ . If you choose the radius $\rho$ for $y$ and $z$ small enough so that the corresponding disks do not contain $x_1$, you'll get a bad configuration.
Indeed, an attempt to choose $x$ or $y$ as one of the centers results in the exclusion of all the centers $x_j$, after which covering $x_1$ gets impossible.
Out of $x_i$, we can choose only one (if $i<j$, then $x_i\in D(x_j,r_j)$). But then, if we choose $x_i$, the point $x_{i+1}$ is not covered.
Now the sequences. Start with any $x_1$ very close to $y$ and not on the line $yz$ so that $d(x_1,z)<1$. Assume that $x_1,x_2,\dots x_j$ and $r_1,\dots,r_{j-1}$ are already constructed and, say $j$ is odd, so $x_j$ is close to $y$. Then the circles centered at $x_j$ and $x$ containing $z$ cross at an angle, so $D(y,1)\setminus \bar D(x_j,d(x_j,z))$ is an open set containing points arbitrarily close to $z$. Choose $x_{j+1}$ to be any point in that difference that doesn't lie on the line $yz$  and satisfies $d(y,x_{j+1})\ge 1-d(z,x_{j+1})>\max_{i\le j}d(z,x_i)+d(z,x_{j+1})\ge \max_{i\le j}d(x_i,x_{j+1})$ and choose $r_j$ anywhere between $d(x_j,z)$ and $d(x_j,x_{j+1})$.
Clearly, we can keep $x_j$ with odd indices at the distance $<1/3$ to $y$ and converging to $y$  and similarly for even indices and $z$.

A: I haven't thoroughly checked the following, so it's a community wiki.
Suppose $K$ is compact, and suppose $r_{x}$ is lower semicontinuous, bounded above, and bounded away from zero:
$$
0 < m \leq r_{x} \leq M,
$$
and suppose $M$ is the least upper bound.  Therefore, this bound is attained at some $x_{M} \in K$.  (Note: I do not require $r_{x}$ to be continuous.  The attention of the bound follows from other considerations.)
Accordingly, one has $r_{x_{M}} = M$.
Let $x_{0} = x_{M}$ and let
$$
K_{1} = K \setminus B(x_{0}, r_{x_{0}}),
$$
where
$$
B(x_{0}, r_{x_{0}}) = B(x_{M}, M).
$$
The restriction of $r_{x}$ to the compact set $\overline{K_{1}}$ is bounded above, hence attains its least upper bound on some $x_{1} \in \overline{K_{1}}$:
$$
r_{x_{1}} = \sup \; \left\{ r_{x} \; : \; x \in \overline{K_{1}} \right\} \leq r_{x_{0}}.
$$
By construction, $x_{1}$ is not in $B(x_{0}, r_{x_{0}})$, and $x_{0}$ is not in $B(x_{1}, r_{x_{1}})$.
We next put
$$
K_{2} = K_{1} \setminus B(x_{1}, r_{x_{1}})
$$
and pick such an 
$$
x_{2} \in \overline{K_{2}}
$$
and which $r_{x}$ attains its least upper bound on $\overline{K_{2}}$.  Note that $B(x_{2}, r_{x_{2}})$ contains neither of $x_{1}, x_{2}$, and that $x_{2}$ is not in either of $B(x_{0}, r_{x_{0}}), B(x_{1}, r_{x_{1}})$.  So, of the three balls constructed so far, none contains the center of another.
After at most transfinitely many repetitions of this step, we get a (possibly finite) sequence $B(x_{i}, r_{x_{i}})$ of balls where none of the balls contains the center of another.  These balls constitute a cover of $K$.  Extracting, if necessary, a finite subcover, we obtain a collection of balls with the desired property.
