Subsets of $\mathbb{R^2}$ using the discrete metric are closed, why? I understand why all subsets of say, $\mathbb{R^2}$ are open with respect to the discrete metric - Let $U$ be a subset of $\mathbb{R^2}$. For all $x \in U$ we can choose an 0 > r > 1 such that we will have an open ball, $B$ that only contains a single point, x itself. And this open ball will be contained entirely in $U$ as $x \in U$ so $U$ is open.
But as we are talking about the discrete metric $U$ is also closed. Closed afaik means that the set contains it's limit points. So is this correct - The limit point(s) of $B$, which only contains a single point $x$, is just that $x$. Hence it's closed? 

In a space with the discrete topology no set has any limit points, so every set vacuously contains all of its limit points. Remember, $p$ is a limit point of a set $A$ if and only if every open neighborhood of $p$ contains a point of $A$ different from $p$. But in a discrete space $\{p\}$ is an open neighborhood of $p$, and obviously it doesn’t contain any point of $A$ different from $p$ no matter what set $A$ is. 
To put it a little differently, let $A$ be any set in a discrete space. In order for $A$ not to be closed, $A$ would have to have a limit point $p$ such that $p\notin A$. But $A$ has no limit points, so it certainly has no limit points that aren’t in it!
It’s probably easier in this context to think of closed sets as the sets that are complements of open sets. In a discrete space every set is open, so every set is the complement of an open set and therefore is also closed. For short we say that every set is clopen, meaning closed and open.
