I am having trouble understanding a small part from theorem 3.2 ($1\implies 2$) of this paper (Between compactness and completeness, G. Beer).
Prerequisite: (1) A sequence $(x_n)$ in a metric space $(X,d)$ called cofinally Cauchy if for $\epsilon>0~\exists~\mathbb N_0\text{ (infinite)}\subset\mathbb N$ such that $d(x_i,x_j)<\epsilon~\forall~i,j\in\mathbb N_0.$
(2) A metric space $X$ is cofinally complete if every cofinally Cauchy sequence in $X$ clusters.
(3) Let $X$ be a metric space and $x\in X.$ If $x$ has a compact neighbourhood set, $v(x)=\sup\{\epsilon>0:S_\epsilon[x]\text{ is compact}\},$ otherwise set $v(x)=0.$
Main theorem 3.2: (1) X is cofinally complete $\implies$ (2) Whenever $(x_n)$ is a sequence in $X$ with $\lim_{n\to\infty}v(x_n) = 0,$ then $(x_n)$ has a cluster point:
Suppose condition (2) fails, and let $(x_n)$ be a sequence in $X$ with $\lim_{n\to\infty}v(x_n) = 0$ that has no cluster point. Without loss of generality, we can assume by passing to a subsequence that $ν(x_n) < 1/n$ and so we can find a sequence $(w^n_j)$ in $S_{1/n}[x_n]$ with no cluster point. Partition $\mathbb N$ into an countable family of infinite subsets $\{K_n: n\in\mathbb N\}.$ Then the assignment $y_j = w^j_n$ for $j\in K_n$ defines a cofinally Cauchy sequence $(y_j)$ in $X$ without a cluster point. Thus (1) fails.
I have verified the theorem other than the fact that $(y_j)$ is cofinally Cauchy. Please tell me why $(y_j)$ is cofinally Cauchy?
Thank you in advance.
EDIT: This is the original version of the answer, which was posted based on the original revision of the question where we had $y_j=w^j_n$. See also some comments below this answer.
From the construction of the sequence $(y_j)$ you can see, that if $j\in K_n$ then you have $y_j\in S_{1/n}[x_n]$. So you have $$d(y_i,y_j)<\frac 2n$$ for any $i,j\in K_n$. Since each $K_n$ is infinite (and thus cofinal in $\mathbb N$), you get that $y_j$ is cofinally Cauchy.