How is $\overline{c_{\epsilon}}:=\overline{\bigcup_{n\geq 1}\left(\bigcap_{k=n}^{\infty}B(x_k,r+\epsilon)\cap K \right)}$ convex in a CAT$(0)$ space? Let $(X,d)$ be a CAT$(0)$ space, $\{x_n\}\subset X$ be bounded and $K\subset X$ be closed and convex. Define $\varphi:\,X\longrightarrow \mathbb{R},$ by $\varphi(x)=\limsup\limits_{n\to\infty}d(x,x_n)$ for each $x\in X.$ Then, there exists a unique point $u\in K$ such that $$\varphi(u)=\inf\limits_{x\in K}\varphi(x).$$
Proof
Let $r=\inf\limits_{x\in K}\varphi(x)$ and $\epsilon>0.$ Then, there exists $x_0\in K$ such that $\varphi(x_0)<r+\epsilon.$ This implies that there exists $N\in \mathbb{N}$ such that 
$$x_0\in\bigcup_{n\geq N}\left(\bigcap_{k=n}^{\infty}B(x_k,r+\epsilon)\cap K  \right)\subset\bigcup_{n\geq 1}\left(\bigcap_{k=n}^{\infty}B(x_k,r+\epsilon)\cap K  \right).$$
Define $c_{\epsilon}:=\bigcup_{n\geq 1}\left(\bigcap_{k=n}^{\infty}B(x_k,r+\epsilon)\cap K  \right).$ It is clear that $c_{\epsilon}$ is convex.
Question: How is $\overline{c_{\epsilon}}$ convex and $c:=\bigcap_{\epsilon>0}\overline{c_{\epsilon}}\neq \emptyset?$
Details: The current paper I'm reviewing is Dhompongsa et al. To show that $\overline{c_{\epsilon}}$ is convex, Dhompongsa et al., referred to  Proposition 1.4(1). I'm just finding it hard to comprehend. Any help, please?
 A: Update : i) Helly Theorem in ${\rm CBA}\ [0]$ :
When $R_i$ is a closed
bounded and convex s.t. $\bigcap\ R_i=\emptyset$, then there is $N$ with $\bigcap_{i=1}^N
  \ R_i=\emptyset$
ii) Each ball in ${\rm CAT}\ [0]$-space is convex
And a closure of ${\rm CAT}\ [0]$ space is ${\rm CBA}\ [0]$-space
(cf ${\rm CBA}$ space needs a complete metric space condition on
${\rm CAT}$, more.)
iii) When $$C_n = \bigcap_{k\geq n}\
\underbrace{B(x_k,r+\epsilon)\bigcap K}_{convex}
$$
then each $C_n$ is convex and $C_n\subset C_{n+1}$ which is an increasing seq
Define $c_\varepsilon =\bigcup_n\ C_n$ which is still
convex since
$C_n$ is increasing sequence
iv) When $x={\rm lim\ inf}\ x_i$, then generally $
\overline{x}\neq {\rm lim\ \inf}\ \overline{x}_i$. But here we
consider convex sets. Hence this may not be a problem.
We will prove by a contradiction Assume that there
is $N$ with $\bigcap_{i=1}^N
  \ c_{1/i}=\emptyset$
By definition of $r$, there is $k\in K$ s.t. $ \phi(k)<r+\epsilon$
By definition of $\phi$ (which is defined through sup), there
is $M$ s.t. $ |x_n-k|\leq r+\epsilon$ for $n>M$.
Hence $c_\epsilon $ contains $k$
Further, for $\epsilon/2$ there is $k'\in K$ s.t. $
\phi(k')<r+\epsilon/2$
And $|x_n-k'|\leq r+ \epsilon/2 $ for $n>M'$
Hence $c_{\epsilon /2}$ contains $k'$ Here for $n>M,\ M'$ $
|x_n-k'|,\ |x_n-k|< r+ \epsilon$ so that $c_{\epsilon/2}\subset
c_\epsilon$

OLD : (1) Each ball is convex so that $c^n_\varepsilon =\bigcap_{k\geq n}\
B(x_k,r+\varepsilon )\cap K$ is convex. Note that
$c_\varepsilon^n\subset c_\varepsilon^{n+1}$ so that $c_\varepsilon$
is convex.
(2) Helly Theorem :
When $B_i$ is a closed bounded and convex and $
\bigcap_i\ B_i=\emptyset$, then there is finite many $B_{n_i}$ s.t.
$ \bigcap_{i=1}^{i_0}\ B_{n_i} = \emptyset$
(cf Alexandrov geometry - Alexander, Kapovitch, and Petrunin, 2016, 142
p.)
(3) When $A=\{a_n\}$, since $
 r=\inf_{x\in K}\ \phi(x)$, then there is $k_i\in K$ s.t.
$$ |r- \phi(k_i)| \leq r+\frac{1}{i} $$
Then there is $n_i$ s.t. $$ n\geq n_i \Rightarrow  |k_i-a_n| \leq
r+\frac{2}{i}
$$
Proof : If not, then there is infinitely many $n$ s.t.
$|k_i-a_n|>
r+\frac{2}{i} $. That is, $\phi(k_i)\geq r+\frac{2}{i}$. Contradiction.
Corollary : $\bigcap_{n\geq n_i}\ B(a_n,r+\frac{2}{i})$
contains $k_i$. That is, $k_i\in c_{\frac{2}{i}}$
Corollary : $\bigcap_\varepsilon\ c_\varepsilon$ is not empty
Proof : $n\geq \max\{ n_{i},n_j\},\ i<j$ implies that $$|a_n-k_i|\leq r+
\frac{2}{i}, \ |a_n-k_j|\leq r+\frac{2}{j} $$
so that $$|a_n-k_i|,\ |a_n-k_j|\leq r+ \frac{2}{i}$$
That is, $k_i,\ k_j\in c_\frac{2}{i}$ That is, any finite
intersection of $c_{\frac{2}{i}}$ is not empty. By Helly theorem, we complete the proof.
