Why discrete set must be countable? I want to show a set, which every point of it is an isolated point. Then this set must be countable. How to show it?
I find this from Wikipedia's article Isolated point, but I don't understand:

A set that is made up only of isolated points is called a discrete set (see also discrete space). Any discrete subset S of Euclidean space must be countable, since the isolation of each of its points together with the fact that rationals are dense in the reals means that the points of S may be mapped into a set of points with rational coordinates, of which there are only countably many.

 A: *

*Because $S$ is discrete, we can find a set $T$ of open intervals such that each point of $S$ is in exactly one of the intervals, and no two intervals intersect.

*For each interval in $T$ we can pick one rational that is contained in it.  Call this set of rationals $R$.

*Because the elements of $T$ are pairwise disjoint, there is a one-to-one correspondence between $T$ and $R$, so $|T|=|R|$.

*$R$ is a subset of the the rationals, so is at most countable; therefore $T$ is at most countable.

*Elements of $T$ are in one-to-one correspondence with elements of $S$, so $|S|=|T|$.

The important thing to think about here is: what part of the proof fails when $S$ is not discrete?  Say, when $S = \{1, \frac12, \frac13, \frac14, \ldots, 0\}$.
A: Intuitively, $x$ is an isolated point of $S$ if we can "draw some circle" around $x$ such that $x$ is the only point in that circle which is an element of $S$. This is because a point is isolated iff it is not a limit point, and $x$ is a limit point of $S$ iff we can find elements of $S$ (other than $x$ itself) arbitrarily close to $x$.
OK, so now suppose every element of $S$ is an isolated point of $S$. Imagine drawing a whole bunch of "isolating bubbles" around the elements of $S$. One thing we can try ot do is, for each $x\in S$, pick a rational $q_x$ in the "isolating bubble" around $x$. The idea then is that since no two elements of $S$ lie in the same "isolating bubble," the rationals we pick should be different.
However, this doesn't quite work. What if the "isolating bubbles" themselves overlap? E.g. consider $S=\{0,1\}$. Then the interval $(-{2\over 3},{2\over 3})$ is an "isolating bubble" around $0$, the interval $({1\over 3}, {4\over 3})$ is an "isolating bubble" around $1$, but the rational number ${1\over 2}$ lies in both of those "isolating bubbles" so we might pick it for $0$ and $1$ simulatenusly.
What we need to do is pick really small "isolating bubbles" - so small that they're guaranteed to not overlap at all. In the case above, we could for example use $(-{1\over 2}, {1\over 2})$ as the "bubble" around $0$ and $({1\over 2},{3\over 2})$ as the "bubble" around $1$. One way to do this is to go back to the original "bubbles" we drew and shrink them a bit:

*

*Since $S$ is isolated, for each $x\in S$ we can find a $\delta_x>0$ such that $(x-\delta_x,x+\delta_x)\cap S=\emptyset$.


*Now let $I_x$ be the smaller interval $(x-{\delta_x\over 2},x+{\delta_x\over 2})$. Show that for $x,y\in S$ distinct we have $I_x\cap I_y=\emptyset$. So we've found isolating bubbles which don't overlap at all.


*Now apply the density of $\mathbb{Q}$: for each $x\in S$ pick some rational number $q_x\in I_x$. Show that the map $S\rightarrow\mathbb{Q}: x\mapsto q_x$ is injective and so $S$ is countable (this is where the non-overlapping-ness of the $I$s comes in).
A: The tag "real-analysis" is added so I presume this is a discrete subset of the reals.
Let $x\in S$ be arbitrary. Then, by the hypothesis, we can choose $\delta(x)$ such that $S\cap(x-\delta(x),x+\delta(x))=\emptyset$ (i.e. isolated). "The rationals are dense in the reals" means that in any open set of reals, we can find a rational in it. For example $(\pi - \frac{1}{n} ,\pi+\frac{1}{n})$ contains a rational for all naturals $n$. Now choose a rational $r(x)=\frac{p}{q}$ in $(x-\delta(x),x+\delta(x))$ such that $p+q$ is minimal(the smallest in this set). If there are multiple of these such that $p+q$ is minimal, choose the one with the smallest $p$. This awkward choice of minimal is to avoid using AoC, we could otherwise just say "choose a rational in the neighbourhood". Now define $f$ to be a function such that $f:S\rightarrow \mathbb{Q}$, $f(x)=r(x)$. By nature of isolated points, this function is injective, thus $|S|\sim |\mathbb{Q}|\sim |\mathbb{N}|$
