Does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms? Given a metric space $(X,d),$ let us call a sequence $(x_n)$ in $X$ to be pseudo-Cauchy if $\lim\limits_{n\to\infty}d(x_n,x_{n+1})=0.$
For example, the sequence $\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)_n$ is pseudo-Cauchy in $\mathbb R$ without being Cauchy.
Now I would like to ask the following question:

Given a pseudo-Cauchy sequence $(x_n)$ in a metric space $(X,d)$ having no constant subsequence, does there always exist a pseudo-Cauchy subsequence of $(x_n)$ having distinct terms?

Intuitively it is appearing to me that there should not always exist such a subsequence; however I failed to construct a counterexample or give a proof.
Please help me!
 A: Such a subsequence always exists. 
No constant subsequence means that each value $v$ coming up in the sequence $\{x_n\}_1^\infty$ has a maximal index $m(v)$ where it occurs.
Let's define the indizes of the subsequence as follows:
$s_1=1, \forall k \ge 1: s_{k+1}=m(x_{s_k})+1.$
That sequence is monotonically increasing (because $\forall n: m(x_n) \ge n$ by definition) and all values $x_{s_n}$ are different, because the next index after $s_k$ is chosen such that all previous values are guaranteed not to come up any more in $\{x_n\}_{s_{k+1}}^\infty$.
It remains to show that $\{x_{s_n}\}_1^\infty$ is pseudo-Cauchy.
So given any $\epsilon > 0$, we know there exists an $N \in \mathbb N$ with 
$$d(x_n,x_{n+1}) < \epsilon, \forall n \ge N, \tag{1}$$
because $\{x_n\}_1^\infty$ is pseudo-Cauchy. Let $K$ be the smallest integer such that $s_K \ge N$.
For any $k \ge K$ we have, recalling that $x_{s_k}=x_{m(x_{s_k})}$ and the definition of $s_{k+1}$:
$$d(x_{s_k},x_{s_{k+1}}) = d(x_{m(x_{s_k})},x_{s_{k+1}}) = d(x_{m(x_{s_k})},x_{m(x_{s_k})+1}) < \epsilon,$$
because $m(x_{s_k}) \ge s_k \ge s_K \ge N$ and $(1)$ above.
Since $\epsilon$ was arbitrary, this means $\lim_{k \to \infty} d(x_{s_k},x_{s_{k+1}}) = 0$, so $\{x_{s_n}\}_1^\infty$ is now proven to be pseudo-Cauchy.
