Q:

How do I show that the set of limit points of $\{x_n\}$ is connected?

Show that if $\{x_n\}$ is a sequence of real numbers such that $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$, then the set of limit of points of $\{x_n\}$ is connected, that is either empty, a single point, or an interval.
My attempt:  Since, $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$ $\implies$ For given $\epsilon>0$, there exist a positive integer $N$ such that for $n>N$,
$|x_{n+1}-x_n|<\epsilon$ $\implies$ $|x_{n+1}|-|x_{n}| \leq |x_{n+1}-x_n|<\epsilon$.
This means for $n>N$, $|x_{n+1}|\leq |x_n|+\epsilon.$
Case 1: If $\{x_{n}\}$ is convergent, say it converges to $x$, and $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=x-x=0$. We know that if $\{x_n\}$ converges, then its limit is unique. So, the set of limit points contain a single element.
Case 2: If $\{x_{n+1}\}$ is divergent, then there is a $M$ and $N$ such that $x_{n+1}>M$ for every $n>N$, $\implies$ $x_{n}$ is divergent(by comparison test). In this case, the set of limit points is empty because sequence is divergent.
Case 3: If $x_{n}=k+\frac{1}{n}$ for $k\in \mathbb{R}$, then $\lim_{n \rightarrow \infty}(x_{n+1}-x_{n})=0$, and every point of $\mathbb{R}$ is a limit point of $x_{n}$. So the set of limit points are the interval.
This shows that the set of limit point of a sequence $x_n$ is connected.
Is this solution correct for the given question?

A:

Either the sequence has no limit points.
Or it has one. In this case, the sequence converges to it.
Otherwise, let $a,b$ be any two limit points (we usually use the term 'cluster points') of $(x_n)$, and let $a<c<b$ and $\epsilon>0$. Without loss of generality, ensure $\epsilon<\min(b-c,c-a)$. There exists $N\in\mathbb{N}$ such that $$n\ge N\implies |x_{n+1}-x_n|<\epsilon\tag{1}$$
Since $a,b$ are limit points, there also exist $N_b>N_a\ge N$ such that $$|x_{N_a}-a|<\epsilon,\qquad|x_{N_b}-b|<\epsilon$$
Now consider the sequence $x_n$ for $N_a\le n\le N_b$. It goes from nearby $a$ to nearby $b$ using steps smaller than $\epsilon$. There must be a value of $n$ in this range such that $|x_n-c|<\epsilon$ otherwise one can split the $n$ into those that give $x_n\le c-\epsilon$ and those that give $x_n\ge c+\epsilon$. The largest $n$ in the first group is not $N_b$ and has $n+1$ in the second group, so $$x_{n+1}-x_n\ge(c+\epsilon)-(c-\epsilon)=2\epsilon$$ contradicting (1).
Since the set of limit points $L$ satisfies $a,b\in L\implies [a,b]\subseteq L$, then $L$ must be an interval itself.

