How can an isolated point be an open set? I have the following definition:

In a metric space $(X,d)$ an element $x \in X$ is called isolated if $\{x\}\subset$ X is an open subset

But how can $\{x\}$ be an open subset? There has to exist an open ball with positive radius centered at $x$ and at the same time this open ball has to be a subset of $\{x\}$ but how can this be if there is only one element?
I'm trying to wrap my head around this, but I can't figure it out. It doesn't make sense for metrics on $\mathbb{R}^n$ since each open ball with some positive radius has to contain other members of $\mathbb{R}^n$.
The only thing I could think of was that we have some $x$ with 'nothing' around it and an open ball that contains only $x$ and 'nothing' (even though a positive radius doesn't make sense since there is nothing), so therefore the open ball is contained in $\{x\}$. But I'm not even sure we can define such a metric space, let alone define an open ball with positive radius containing only $x$ and 'nothing'.

Take, for instance, your set to be the integers, with the metric inherited from the metric on $\mathbb{R}^1$. Then any singleton is open, and so every point is isolated. This is because, as you said, these points have 'nothing' around them (if you look close enough).
