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?