I have been reading George M. Bergman's supplementary notes and exercises to Rudin's Introduction to mathematical analysis (which can be found here https://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf), and I am stuck in problem 3.2:6 part b), which states the following:
Let $X$ be a metric space and $(s_n)$ a sequence in $X$ such that $d(s_n,s_{n+1}) \rightarrow 0$. Let $E$ denote the subsequential limit set of $(s_n)$ (that is, $p\in E$ whenever there is a subsequence of $(s_n)$ which converges to $p$). Show that for every $\epsilon > 0$, all but finitely many natural numbers $n$ have the property that there exists $p \in E$ such that $d(s_n,p)< \epsilon$. (I assume that one must suppose that $E$ is not empty, otherwise the proposition would be false).
Here is what I've tried:
If $S$ denotes the range of the sequence $(s_n)$, then we can split $S$ in two disjoint subsets: for some $\epsilon > 0$, define $A : = \{s_n \in S | \forall p \in E: d(s_n, p) \geq \epsilon\}$ and $B := \{s_n \in S | \exists p \in E: d(s_n, p) < \epsilon \}$. Clearly both sets are disjoint, $S = A \cup B$, and $B$ has infinitely many elements. I've tried to prove the statement by asumming that $A$ has infinitely many elements, and then conclude that $\lim_{n\rightarrow\infty} d(s_n, s_{n+1}) \neq 0$, or assuming that $\lim_{n\rightarrow\infty} d(s_n, s_{n+1}) = 0$ and then obtain a contradiction, but I see no way of getting to the desired result. Maybe the proposition is false, but I've not been able to figure out a counterexample.
I would be really grateful if you could help me to solve this problem, not with the complete answer, but with some hint, suggestion, or material that could be related to this.