Bounded-ness of range We are give a sequence of the form $x_n = \frac{1}{n}$ in the complex metric space.
This sequence of course has a limit at $0$, the range is clearly infinite, however, it is said that the sequence is bounded.
Now, a sequence is bounded if its range is bounded, or if it is given to be both bounded from above and bounded from below.
Why is the sequence bounded from above?
 A: A sequence in a set $X$  is the same thing as a function from the set $\mathbb{N}$ of natural numbers to $X$.
For example, if $a_n$ is a sequence in the set $X$, then we can define a function $A:\mathbb{N}\to X$ by
$$A(n)=a_n.$$
Conversely, given a function $A:\mathbb{N}\to X$, we can define the sequence $a_n$ by the above formula as well. Thus, sequences and functions from $\mathbb{N}$ are equivalent notions.

The "range" of a sequence is just the range of the corresponding function. That is, if $a_n$ is a sequence in the set $X$, then the range of the sequence is
$$\{a_n\in X: n\in\mathbb{N}\}=\{A(n):n\in\mathbb{N}\}=A(\mathbb{N}).$$
Clearly, this set is infinite precisely when there are infinitely many distinct elements of $X$ that are elements of the sequence $a_n$. Thus, it is true that the sequence $x_n=\frac{1}{n}$ in $\mathbb{C}$ has an infinite range.
However, this is separate from the notion of a sequence being bounded. 
Given a metric space $M$, we say that a set $S\subseteq M$ is "bounded" when $S$ is contained inside some ball of the metric space. That is, $S$ is bounded when there is some $p\in M$ and real number $r>0$ such that
$$S\subseteq B(p,r)=\{q\in M: d(p,q)\leq r\}.$$
Given a set $Y$ and a metric space $M$, we say that a function $f:Y\to M$ is bounded when the range of the function, namely the set
$$f(Y)=\{m\in M: m=f(y)\text{ for some }y\in Y\},$$
is a bounded set.
Now, we simply apply this notion to the function corresponding to a sequence. A sequence $a_n$ in a metric space $M$ is equivalent to the function $A:\mathbb{N}\to M$ defined by $A(n)=a_n$. Then a sequence is bounded when the range of the sequence is a bounded set.
We can see that this is the case for your example, because we have
$$x_n=\frac{1}{n}\in B(0,1)=\{z\in\mathbb{C}: d(z,0)\leq 1\}=\{z\in\mathbb{C}: |z|\leq 1\}$$
for all $n$. 
