Properties of Sequences with diminishing steps Suppose $x_n$ is a sequence in $\mathbb{R}^N$ that satisfies
$$
\|x_{n} - x_{n-1}\|\to 0.
$$
Assume $C$ is the set of limit points (a.k.a. cluster points) of $x_n$ and assume $C$ is nonempty. Can it be shown that
$$
d(x_n,C)\to 0?
$$
Where $d(x,C)=\min\{\|y-x\|:y\in C\}$. (Note that the set of limit points is always closed so this number is well-defined).
 A: This is not true. For example, let $f : [0,\infty) \to \mathbb C$ be defined by 
$$f(t) = \begin{cases} 1 - e^{2\pi i t} & \text{when } t\in [0,1]  \\ 2 - 2 e^{\pi i (t-1)} & \text{when } t\in [1,3] \\ 3- 3 e^{\frac{2\pi i (t-3)}{3}} & \text{when } t\in[3,6]\\\vdots & \vdots\end{cases}$$  
That is, the function goes around the circle $C_1$ of radius $1$ centered at $1$, then the circle $C_2$ of radius $2$ centered at $2$ and so on, note that $f$ is chosen such that the speed is always $2\pi$ (note that it holds when $t\notin 1,3,6,\cdots$). 
Now define $x_n = f(s_n)$, where $s_n =1 + \frac 12 + \cdots + \frac 1n$. Then 
$$\|x_n- x_{n+1} \| = \left\| \int_{s_n}^{s_{n+1}} f(t) dt\right\|\le \int_{s_n}^{s_{n+1}} |f'(t)| dt = \frac{2\pi}{n+1}\to 0$$
Claim: $C$ is the imaginary axis. 
Obviously $C \subseteq \{ z\in \mathbb C :\text{Re} z \ge 0\}$. If $\text{Re} z>0$, then $z$ is inside a circle $C_N$ of radius $N$ centered at $N$ when $N$ is large enough. Since there are only finitely many $x_n$ in this circle, this shows that $z\notin C$. Thus $C$ lies in the imaginary axis. 
Now let $it$, where $t\in \mathbb R$ and $\epsilon >0$ be arbitrary. Then there is $N$ so that $C_n \cap \text{Box}_\epsilon (it) \neq \emptyset$ for all $n\ge N$, where $\text{Box}_{\epsilon} (it)$ is the square centered at $it$ with side length $\epsilon$. Let $a_n, b_n\in \mathbb R$ so that $f[a_n,b_n] \subseteq \text{Box}_\epsilon (it) \cap C_n$ and $a_n, b_n$ lie in the lower and upper edge of $\text{Box}_\epsilon (it)$ respectively. Then $\|f(a_N) - f(b_N)\| \ge \epsilon$. This implies $|a_n - b_n|\ge \frac{\epsilon}{2\pi}$. Since $a_n, b_n \to \infty$ as $n\to \infty$, there are $k\in \mathbb N, n\ge N$ so that $s_k  \in [a_n, b_n]$ and thus $x_k = f(s_k)$ satisfies $\| it - x_k\|\le \frac{\epsilon}{\sqrt 2}$. Thus $it \in C$. Since $t$ is arbitrary, the claim is proved. 
Note that $d(x_n, C)$ does not converge to $0$. 
Remark: If $\{x_n\}$ is additionally bounded, then the assertion is true, even without the assumption on $\|x_n - x_{n+1}\|$: Assume the contrary that $d(x_n,C)$ does not converge to zero. Then there is a subsequence $\{x_{n_k}\}$ so that 
$$\tag{1} d(x_{n_k},C) \ge \epsilon_0$$
 for some $\epsilon_0>0$. Since $\{x_n\}$ is bounded, a subsequence of $\{x_{n_k}\}$ (which we still call $\{x_{n_k}\}$) converges to some $c$. By definition $c\in C$ and so $d(x_{n_k},C)\le d(x_{n_k} ,c) \to 0$, a contradiction to $(1)$. 
A: Let me offer this as an addendum to @ArcticChar's answer. It's the same kind of example in $\mathbb C = \mathbb R^2,$ but might be simpler. I'll use complex number notation; it will be good to draw a picture to visualize the sequence.
We start with steps of $1:$ $0,1,1+i,1,0.$
Next, continue with steps of $1/2$: $0,1/2,1,3/2,2,2+i/2,2+i,2+i/2, 2,3/2,1,1/2,0.$
Follow with steps of $1/3$: $0,1/3,2/3,1,4/3,5/3,2,7/3,8/3,3, 3+i/3,3+2i/3 ,3+i,$ then retrace in steps of $1/3$ back to $0.$
Continue in this way to obtain a sequence $x_n$ in $\mathbb C$ such that $|x_{n+1}-x_n|\to 0,$ whose range contains all rational numbers in $[0,\infty),$ and such that $x_n$ has no cluster points in $\mathbb C\setminus [0,\infty).$ Because the range of $x_n$ contains all rationals in $[0,\infty),$ we have $C = \mathbb R.$ We also have the points $\{m+i: m \in \mathbb N\}$ in the range of $x_n$, showing $d(x_{n_k}, C) = 1$ along a subsequence, negating $d(x_n,C) \to 0.$
