Discrete Sets in Certain Metric Spaces are Countable I've recently started my first real analysis course, and we're studying metric spaces from Rudin's Principles of Mathematical Analysis (Baby Rudin).
We have the following definition: 
Let $(X,d)$ be a metric space, and $A \subset X$. A point $p \in A$ is said to be isolated if $p$ is not a limit point of $A$. Furthermore, $A$ is said to be discrete if all its points are isolated. 
Now, consider $\mathbb{R}^{n}$ with the usual metric $d(x,y)=|x-y|$. Prove that if $A \subset \mathbb{R}^{n}$ is discrete, then it is at most countable. 
What I had in mind was something along the lines of:
Since we know that for every $p \in A$ there exist some $r \in \mathbb{R}$ such that $B_{r}(p)  \cap A = \left \{ p \right \}$. We can find some rational $q$ with $0<q<r$, and isolate $p$ with the ball of radius $q$. Then we define a function which assings to each $p$ said $q$. If said function were injective, we would be done, but I don't know how to do that, since all points in $A$ could very well be isolated by finitely many different reals $r$.
Any ideas?
Thanks in advance!
PS: I've previously shown that a discrete set $A \subset \mathbb{R}^{n}$ is compact if and only if it is finite. But it could not think of any useful ways to apply this to the problem at hand. 
 A: Hint. Show that for each point $p\in A$ there is an open ball $B$ with rational center and radius, such that $B\cap A=\{p\}$. (I.e., the radius is a rational number, the center is a point in $\mathbb R^n$ all of whose coordinates are rational numbers.) The set of all such balls is countable.
A: In reference to the hint,we do the following. 
Choose  r positive real such that B:= B(p,r)'s closure, has no point of A other than p.let (r(n)) be a sequence of rational coordinates in B(p,r)such that it converges to p.Let s(n)=d(r(n),bdB) and let p(n)=d(r(n),p) for all n. p(n) converges to 0 and s(n) converges to r.
Let  v(n)=s(n)-2p(n) for all n.Thus v(n) converges to r.Since r is positive,there exists N a natural number,
Such that v(N)=s(N)-2p(N)is positive.Hence 2p(N) is strictly less than s(N).
B(r(N),2p(N)) is contained in B(p,r) since Eucledian balls are convex and straight lines in Eucledian spaces are connected.
Now there exists a positive rational t such that t is in between p(N) and 2p(N). Also r(N) is rational in all coordinates.Hence B(r(N),t) is inside B thus B(r(N),t) has p as the only point of A. 
(Note p is in B(r(N),t)). Hence is the proof.
