Discrete metric means all sets are countable? I was working on a proof of "Show that if $A \subseteq \Re^2$ is discrete, then A is a countable set." and I thought about using the discrete metric 
($d(x,y)=\delta_{xy}$) on the set as an example metric, just to get some intuitive understanding. 
However, using this metric on the whole space $  \Re^2$ means $  \Re^2$ is discrete, because there exists an open ball of radius ½ around every point. So $ \Re^2$  is countable, which is decidedly untrue. 
While this claim then might be true for other metrics, I don't understand how the choice of metric can change the countability, which is a property of the number of elements, not the topology. 
Thanks!
 A: There is the notion of a discrete topological space, which is a topological space $X$ in which every subset of $X$ is open. The task is to prove that if a subset $S\subseteq\mathbb{R}^2$, when endowed with the subspace topology from the standard topology on $\mathbb{R}^2$, is a discrete topological space, then the set $S$ must be countable.
This is entirely separate from your (correct) observation that one can take any set $X$ and endow it with the discrete metric, no matter the cardinality of $X$.
A: You are confusing the subspace topology on $A$ with the topology on $\Bbb R^2$.  What the exerices is saying is that if the subspace toplogy that $A$ inherits from the standard topology on $\Bbb R^2$ is discrete, then $A$ is countable.  When you introduce the discrete metric on $\Bbb R^2$, you are changing the problem
A: The change of metric does not change the number of points in the space, but it can change how many subsets you can have of some diameter.  The problem assumes the standard Euclidean metric.  You are correct that $\Bbb R^2$ is uncountable, and with the discrete metric every point is isolated, so you have an uncountable set of balls of radius $1/2$.  In the standard metric having $A$ be discrete means there is some minimum Euclidean distance $\epsilon$ between points of $A$.  That means you can draw a ball of radius $\epsilon /2$ around each point of $A$ and all the balls will be disjoint.  Now you can show that there are only countably many balls-label each with a point from $\Bbb {Q \times Q}$ that is not in any other ball.
