Suppose $(X,d)$ be a metric space. Let $(a_n)$ be a sequence in $X$ such that $(a_{n})$ has no Cauchy subsequence. Let $A=\{a_{n}:n\in\mathbb{N}\}$, is it true every Cauchy sequence $(b_n)$ in $A$ have a constant tail (In other words, there exist $N \in \mathbb{N}$ such that $b_n=b_N$ for every $n \geq N$)?
I asked this before (see About Cauchy sequence). However, I don't fully understand the answer and I was advised to ask a new question about it.
If $(b_n)$ is a Cauchy sequence in $A$, there are two possibilities. If $(b_n)$ takes only finitely many values, then it must have a constant tail (or else there would be distinct elements of $A$ which appear in $(b_n)$ infinitely many times, in which case they would both be limit points and so $(b_n)$ would not be Cuachy). The other possibility is that it takes infinitely many values. In this case, we can construct a subsequence of $(a_n)$ which is also a subsequence of $(b_n)$. We do this by setting $a_{n_1}=b_1$, then given $a_{n_1},\ldots,a_{n_k}$, we have so far only gone past finitely many elements of $(a_n)$, but there are infinitely many $b_n$, so we choose $a_{n_{k+1}}$ to be further along both sequences than anything we have passed thus far.
Since $(b_n)$ is Cauchy, all subsequences are Cauchy, so we have a Cauchy subsequence of $(a_n)$, a contradiction. Hence every Cauchy sequence in $A$ must have a constant tail.