I have 2 contradictory solutions to the follower problem:
The problem:
Let $X$ be an infinite set. For $p \in X$ and $q\in X$ define $$ d(x,y)= \begin{cases} 0\qquad&\text{if and only if $x=y$}\\\ 1&\text{otherwise} \end{cases} $$ Which subsets are closed and which are open?
My "solution":
Every point is isolated: Consider balls of radius $1/2$ around each x. Which is to say no point is a limit point in $X$ (a limit point will have other points of $X$ in the neighborhood). These points, which are not limit points, will be contained in any subset of $X$. Thus no subset of $X$ will contain limit points. Thus no subset of $X$ will be closed.
But:
Balls of radius $1/2$ centered at $x \in X$ are open because the topology of a metric space has all open balls defined to be open. The complement of this ball is closed by proof in Rudin. It is also non-empty (containing all the other points that are not $x\in X$). Thus the open sets are unions of open balls containing points and their complements are the closed sets.
I produced no closed sets in the first argument and a bunch in the second -- what happened?? After reviewing the "similar questions", I think the second explanation is correct. I don't see why the first one is wrong though.