How would you show that an infinite set $\displaystyle \mathbb{N}$ has no cluster points...
If $\displaystyle n\in\mathbb{N}$ can the open set $\displaystyle \left(n-\frac{1}{2},n+\frac{1}{2}\right)$ contain an point of $\displaystyle \mathbb{N}$ other than $\displaystyle n$?
If $\displaystyle x\notin\mathbb{N}$ is there an open set containing $\displaystyle x$ and no point of $\displaystyle \mathbb{N}$.
Hint: $\displaystyle \left( {\exists j \in \mathbb{Z}} \right)\left[ {j < x < j + 1} \right]$.
Let $\displaystyle \delta = \min \left\{ {\left| {x - j} \right|,\left| {j + 1 - x} \right|} \right\}$
A little more generally, any set $\displaystyle E$ such that $\displaystyle E$ such that $\displaystyle d(e,e')>\delta>0$ for all $\displaystyle e,e'\in E$ has no limit points. To see this let $\displaystyle x$ be an arbitrary point and consider $\displaystyle B_{\frac{\delta}{2}}(x)$. It either contains one or zero points of $\displaystyle E$. If the former we're done, if the latter take the open ball of one-half the distance from $\displaystyle x$ to that point.