Q:

Finding an example of a bounded sequence in a complete metric space such that the sequence has no partial limit

I'm working through an analysis text and I've come across this exercise:
Give an example of a complete metric space $X$ and a bounded sequence $\left(x_{n}\right)$ in $X$ such that the sequence $\left(x_{n}\right)$ has no partial limit. 
The reason that I'm stuck with this one is because I'm shaky about what it means for a sequence to be bounded. Does this mean that for any positive integers $n$ and $m$, the set of all possible $|x_{n}-x_{m}|$ is a bounded set of real numbers?
If so, I'm not sure how to relate the boundedness of a sequence to the completeness of a metric space.
I know that a metric space $X$ is said to be complete if every Cauchy sequence in $X$ converges to a point in $X$... what does this have to do with bounded sequences?!? Thanks in advance for any help.
EDIT: I'm going to add a couple of definitions.
A sequence is said to be bounded if there exists a constant $C\gt 0$ such that for any positive integers $n$ and $m$ the inequality $|x_{n}-x_{m}|\lt C$ holds.
A point $x$ in a metric space $X$ is said to be a partial limit of a sequence $\left(x_{n}\right)$ if for every $\epsilon\gt 0$ there are infinitely many values of $n$ for which $x_{n}\in B\left(x,\epsilon\right)$. 

A:

Fix any point $y_0\in X$, and define a sequence $(x_n)$ to be bounded if there exists a constant $C>0$ such that $d(x_n,y_0)\leq C$ for all $n$. This definition does not depend on $y_0$, since if you choose any other $y_1\in X$, then by the triangle inequality,
$$|d(x_n,y_0)-d(x_n,y_1)|\leq d(y_1,y_0),$$
where the right-hand-side is a constant, so the sequence $(x_n)$ is bounded "with respect to $y_0$" if and only if it is bounded "with respect to $y_1$".
Hint for the exercise: Consider a space with no limit points.

