Let $\{x_{n}\}_{n=1}^{\infty}$ be a sequence that has $0$ and $1$ as subsequential limits. Suppose that for every $n$, $|x_{n+1}-x_{n}| < \frac{1}{10}.$ Prove that the set $PL({x_n})$ - (the set of all Subsequential limits of $x_{n}$) has at least $3$ objects.
I tried to prove it by contradiction, but I can't find a way to show that $PL({x_n})$ must have more than $2$ objects.
Since $|x_{n+1}-x_{n}| < \frac{1}{10}$ and $0$ and $1$ are subsequential limits, there are infinitely many $x_n$'s that belong to $(\frac{2}{10},\frac{8}{10})$. Therefore, there exists $x_{n_k}$ such that for all $k$, $x_{n_k}\in (\frac{2}{10},\frac{8}{10})$. But $(x_{n_k})_k$ is a bounded sequence, thus it has a subsequence $(x_{n_{m_k}})$ that converges to some $L\in [\frac{2}{10},\frac{8}{10}]$.
Besides $(x_{n_{m_k}})$ is subsequence of $(x_n)$.Thus $L\in PL(x_n)$, and $L\notin \{0,1\}$.