A Theorem On Compact Connected Metric Spaces by Stadje I recently came across a surprising theorem, due to Wolfgang Stadje, a special case of which states that:

Let $(X,d)$ be a compact connected metric space. Then there exists a unique real number $c$ such that for any positive integer $n$, and for any $x_1,\ldots,x_n\in X$, there exists $x\in X$ such that $$c=\frac{1}{n}\sum_{i=1}^n d(x,x_i)$$

The original proof of the general statement can be found here.
This proof is unaccessible to me.
Since this proof was published in 1981, it is possible that simpler proofs have been found in subsequent years, especially for the special case mentioned above.
I haven't been able to find any other proof on the internet though.
Does anybody know a more accessible proof of the result mentioned above?
 A: $\newcommand{\mc}{\mathcal}$
$\newcommand{\lrset}[1]{\left\{#1\right\}}$
$\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\R}{\mathbf R}$
We will just show the existence of $c$.
For a fixed $n$ and a tuple $x=(x_1 , \ldots, x_n)\in X^n$, define
$$
\mc I(x) =
\lrset{
\frac{1}{n}\sum_{i=1}^n d(x_i, p):\ p\in X
}
$$
The compactness and connectedness of $X$ implies that $\mc I(x)$ is a closed interval.

Lemma 1. Let $x=(x_1 , \ldots, x_n)$ and $y=(y_1 , \ldots, y_n)$ be points in $X^n$.
  Then $\mc I(x)$ and $\mc I(y)$ have non-empty intersection.

Proof.
Suppose not.
Then without loss of generality assume that $\sup \mc I(x)< \inf \mc I(y)$.
Then for all $p$ and all $q$ in $X$ we have that
$$
\sum_{i=1}^n d(x_i, p)< \sum_{j=1}^n d(y_j, q)
$$
This implies
$$
\sum_{i=1}^n d(x_i, y_k) < \sum_{j=1}^n d(y_j, x_\ell)
$$
for all $1\leq k, \ell\leq n$.
Therefore
$$
\sum_{k=1}^n\sum_{i=1}^n d(x_i, y_k) < \sum_{\ell=1}^n\sum_{j=1}^n d(y_j, x_\ell)
$$
But this is a contradiction since LHS and RHS are equal in the last expression.

Lemma 2.
  Let $x=(x_1 , \ldots, x_m)\in X^m$ and $y=(y_1 , \ldots, y_n)\in X^n$.
  Then $\mc I(x)$ and $\mc I(y)$ have non-empty intersection.

Proof.
Suppose not.
Then one of the following two happen
$$
\frac{1}{m}\sum_{i=1}^m d(x_i, p)< \frac{1}{n}\sum_{j=1}^n d(y_j, q), \quad \forall p, q\in X
$$
or
$$
\frac{1}{n}\sum_{j=1}^n d(y_j, q) < \frac{1}{m}\sum_{i=1}^m d(x_i, p), \quad \forall p, q\in X
$$
Without loss of generality suppose the first happens.
Define the tuples
$$
\tilde x= (x_1 , \ldots, x_1,\ x_2 , \ldots, x_2,\ldots,\ \underbrace{ x_i , \ldots, x_i}_{n \text{ times}},\ \ldots,\ x_m , \ldots, x_m)
$$
$$
\tilde y= (y_1 , \ldots, y_1,\ y_2 , \ldots, y_2,\ldots,\ \underbrace{ y_j , \ldots, y_j}_{m \text{ times}},\ \ldots,\ y_n , \ldots, y_n)
$$
By our assumption, it follows that $\mc I(\tilde x)$ and $\mc I(\tilde y)$ do not intersect.
This contradicts Lemma 1.

Lemma 3.
  Let $\set{\mc I_\alpha}_{\alpha \in J}$ be a collection of closed intervals in $\R$, each contained in $[0, D]$, where $D$ is a positive constant.
  Then the intersection $\bigcap_{\alpha\in J}I_\alpha$ is non-empty.

Proof.
It is easy to prove that every finite subcollection of $\set{I_\alpha}_{\alpha\in J}$ have a non-empty intersection.
Then by the compactness of $[0, D]$ we infer that the desired intersection is also non-empty.
Now we are in a position to finish the proof of the theorem.
By Lemma 2 we have that for any $m$ and $n$ the intervals $\mc I(x)$ and $\mc I(y)$ intersect for any $x\in X^m$ and $y\in X^n$.
Now by Lemma 3 we infer that the intersection $\mc I:=\bigcap_{x\in \mathbf X}I(x)$ is non-empty, where $\mathbf X=\bigcup_{n=1}^\infty X^n$.
Any point $c$ chosen in $\mc I$ satisfies the requirement of Theorem.
