Quasi-Cauchy sequences

For the R-to-L of 1. suppose $(x_n)_n$ is quasi-Cauchy and has a cluster point $p$ but $(x_n)_n$ is not Cauchy. Then $(x_n)_n$ does not converge to $p,$ so there exists $r>0$ such that the set $$S=\{n: |x_n-p|\geq r\}$$ is infinite.

Take $n_0$ such that $n> n_0\implies |x_n-x_{n+1}|<r/3.$ Now the set $$T=\{n:|x_n-p|< r/3\}$$ is also infinite because $p$ is a cluster point of $(x_n)_n$.

For $n_0<n\in T$ there exists $n'> n$ with $n'\in S.$ So for $n_0<n\in T$ let $f(n)$ be the least $n'>n$ such that $n'\in S.$

Observe that the set $U=\{f(n): n_0<n\in T\}$ is infinite . We have $$m\in U\implies (\;|x_m-p|\geq r \;\land \;|x_m-x_{m-1}|<r/3 \;\land\; |x_{m-1}-p|<r\;).$$ So for $m\in U$ we have $$x_m\in V=[p-4r/3,p-2r/3]\;\cup \; [p+2r/3, p+4r/3].$$ But there are infinitely many $m\in U,$ so $(x_n)_n$ must have a cluster point $q\in V$, and obviously $q\ne p.$

Summary: A quasi-Cauchy sequence $(x_n)_n$ with exactly one cluster point $p$ must be convergent to $p$, otherwise $(x_n)_n$ would have another cluster point $q$.

1. Any sequence of real numbers which has exactly one cluster point is a Cauchy.
2. If $(a_n ) $ is not Cauchy sequence then it is not convergent. Hence there exists a subseqences $(x_{n_k } )$, $(y_{m_k} )$ of $(a_n )$ such that $x_{n_k }\to x, y_{n_k }\to y$ and $x\neq y $ and $n_k \neq m_k $ for all $k.$ Consider a subsequence $(z_k)$ of $(a_n)$ defined by $z_{2k-1} =x_{n_k} , z_{2k} =y_{m_k}.$ Then $$\lim_{k\to\infty} |z_k -z_{k+1}|=\lim_{k\to\infty} |x_{n_k} -y_{m_k}|=|x-y|\neq 0$$ hence the subsequence $(z_k)$ is not quasi-Cauchy.