Given that $\underline\lim \limits_{n \to \infty}x_n = A < B = \overline\lim \limits_{n \to \infty}x_n$, and $\lim \limits_{n \to \infty} (x_{n+1} - x_n) = 0$，prove that ${x_n}$'s set of accumulation points is exactly $[A, B]$.
It's easy to find two convergent subsequences of ${x_n}$, which converges to $A$ and $B$ respectively. However, doing so would render $\lim \limits_{n \to \infty} (x_{n+1} - x_n) = 0$, one of the two known conditions, useless.
Fix an element $x \in [A,B]$. Fix an $\epsilon > 0$. Because $x_{n+1} - x_n \to 0$, there exists an $N$ such that $n \geq N \implies |x_{n+1} - x_n| < \epsilon$.
Now, choose an $n_0 > N$ such that $|x_{n_0} - A| < \epsilon$. There exists an $n_1 > n_0$ such that $|x_{n_1} - B| < \epsilon$. Consider the elements $x_{n_0},x_{n_0+1},\dots,x_{n_1 - 1},x_{n_1} = x_{n_0 + K}$. We note that for each $i$, we have $|x_{n_0 + i} - x_{n_0 + i - 1}| < \epsilon$. Note, then, that the interval $(x_{n_0 + i} - \epsilon,x_{n_0+i} + \epsilon)$ contains both $x_{n_0+i-1}$ and $x_{n_0+i+1}$ for $i = 1,\dots,K-1$. Of course, $(x_{n_0} - \epsilon,x_{n_0} + \epsilon)$ contains $A$, and $(x_{n_0 + K} - \epsilon,x_{n_0+K} + \epsilon)$ contains $B$.
Use this to conclude that $$ [A,B] \subset \bigcup_{i=0}^K (x_{n_0 + i} - \epsilon, x_{n_0 + i} + \epsilon) $$ Thus, our arbitrary $x$ from the start satisfies $x \in (x_{n_0 + i} - \epsilon, x_{n_0 + i} + \epsilon)$ for some $i$. That is, $|x - x_{n_0 + i}| < \epsilon$. In other words: for any $x \in [A,B]$ and for any $\epsilon > 0$, there exists an $n$ (in this case $n = n_0 + i$) such that $|x_n - x| < \epsilon$.
So, $x$ is an accumulation point. So, every element of $[A,B]$ is an accumulation point, as desired.