How to prove- If $E \subseteq \mathbb{R}$ does not contain any of its limit point then $E$ is countable Let $ E $ be a subset of $ \mathbb{R} $, the set of all real numbers. If $ E $ does not contain any of its limit points, then how can I prove that $ E $ is countable?
Thank you!
 A: Suppose that $ E $ is uncountable. It suffices to show that $ E $ must contain a limit point. (Think of the contrapositive statement; $ A \to B $ is equivalent to $ \neg B \to \neg A $.)
Assume for the sake of contradiction that $ E $ does not contain any of its limit points. Then for every $ p \in E $, there exists a positive number $ \epsilon_{p} $ such that the interval $ (p - \epsilon_{p},p] $ is disjoint from $ E \setminus \{ p \} $.

Claim $ \{ (p - \epsilon_{p},p] \}_{p \in E} $ is a collection of disjoint non-degenerate intervals of $ \mathbb{R} $.
Proof of the claim: As $ \epsilon_{p} > 0 $, it is clear that $ (p - \epsilon_{p},p] $ is non-degenerate. Next, assume for the sake of contradiction that we have distinct $ p,q \in E $ such that $ (p - \epsilon_{p},p] \cap (q - \epsilon_{q},q] \neq \varnothing $. Without loss of generality, we may suppose that $ p < q $. However, this would imply that $ p \in (q - \epsilon_{q},q] $, which is a contradiction because $ (q - \epsilon_{q},q] \cap (E \setminus \{ q \}) = \varnothing $ by construction. The claim is thus established.

Now, by the claim, we see that $ \{ (p - \epsilon_{p},p] \}_{p \in E} $ is an uncountable collection of disjoint non-degenerate intervals, which is yet another contradiction. Therefore, our initial assumption that $ E $ does not contain any of its limit points is false.
Conclusion $ E $ must contain a limit point. Therefore, if $ E $ does not contain any limit point, it must be a countable set.

Addendum
As kindly pointed out by Jonas in his comments below, this result is true for a separable metric space (or pseudo-metric space). My choice of half-open intervals is applicable only to $ \mathbb{R} $, so for the general case, one can choose an $ \epsilon_{p} > 0 $ for every $ p \in E $ such that $ \mathbb{B}(p;\epsilon_{p}) \cap (E \setminus \{ p \}) = \varnothing $. Then $ \left\{ \mathbb{B} \left( p;\dfrac{1}{2} \epsilon_{p} \right) \right\}_{p \in E} $ is an uncountable collection of disjoint open balls, which would violate separability.
A: Haskell Curry’s answer is more than sufficient, but I thought that it might be instructive to put it into a more general topological context.
More generally, let $X$ be a topological space. To say that $E\subseteq X$ does not contain any of its limit points is to say that $E$ is a discrete subset of $X$: each $x\in E$ has an open nbhd $U_x$ such that $U_x\cap E=\{x\}$.

Definition: The spread of $X$, denoted by $s(X)$, is the smallest infinite cardinal $\kappa$ such that $|D|\le\kappa$ for all discrete $D\subseteq X$. In other words, $$s(X)=\max\big\{\omega,\sup\{|D|:D\subseteq X\text{ is discrete}\}\big\}\;.$$

What you’ve been asked to prove is that $s(\Bbb R)=\omega$.

Definition: A family $\mathscr{N}$ of subsets of $X$ is a net if for each open $U\subseteq X$ there is a family $\mathscr{N}_U\subseteq\mathscr{N}$ such that $U=\bigcup\mathscr{N}_U$. The net weight of $X$, denoted by $nw(X)$, is the smallest infinite cardinal $\kappa$ such that $X$ has a net of cardinality $\kappa$. In other words, $$nw(X)=\max\big\{\omega,\min\{|\mathscr{N}|:\mathscr{N}\text{ is a net for }X\big\}\;.$$ The weight of $X$, denoted by $w(X)$, is defined similarly, but with base replacing net: $$w(X)=\max\big\{\omega,\min\{|\mathscr{N}|:\mathscr{N}\text{ is a base for }X\big\}\;.$$

Clearly any base of the topology of $X$ is a net, so $nw(X)\le w(X)$. In particular, we know that $\Bbb R$ is second countable, so $\omega\le nw(\Bbb R)\le w(\Bbb R)=\omega$, and therefore $nw(\Bbb R)=\omega$.
The point of all this is that it’s not hard to prove in general that $s(X)\le nw(X)$, from which we immediately get $s(\Bbb R)=\omega$, the desired result.

Proposition: For any space $X$, $s(X)\le nw(X)$.
Proof: Let $D$ be a discrete subset of $X$, and let $\mathscr{N}$ be a net for $X$ of cardinality $nw(X)$. Each $x\in D$ has an open nbhd $U_x$ such that $U_x\cap D=\{x\}$. For each $x\in D$ there is an $\mathscr{N}_x\subseteq\mathscr{N}$ such that $U_x=\bigcup\mathscr{N}_x$, so there is an $N_x\in\mathscr{N}_x$ such that $x\in N_x\subseteq U_x$. Clearly $N_x\cap D=\{x\}$ for each $x\in D$, so the map $D\to\mathscr{N}:x\mapsto N_x$ is an injection, and $|D|\le|\mathscr{N}|=nw(X)$. $\dashv$

In fact even stronger statements are possible.

Definition: The density of $X$, denoted by $d(X)$, is the smallest cardinality of a dense subset of $X$ (or $\omega$ if that is finite): $$d(X)=\max\big\{\omega,\min\{|D|:\operatorname{cl}D=X\}\big\}\;.$$ The hereditary density of $X$, denoted by $hd(X)$, is the supremum of the densities of subsets of $X$: $$hd(X)=\sup\{d(Y):Y\subseteq X\}\;.$$ The Lindelöf degree of $X$, denoted by $L(X)$, is the smallest infinite cardinal $\kappa$ such that every open cover of $X$ has a subcover of cardinality at most $\kappa$. The hereditary Lindelöf degree of $X$, denoted by $hL(X)$, is defined similarly to the hereditary density: $$hL(X)=\sup\{L(Y):Y\subseteq X\}\;.$$

Thus, $X$ is separable iff $d(X)=\omega$, $X$ is hereditarily separable iff $hd(X)=\omega$, $X$ is Lindelöf iff $L(X)=\omega$, and $X$ is hereditarily Lindelöf iff $hL(X)=\omega$.
It’s very easy to show that $s(X)\le hd(X),hL(X)$ for any space $X$: if $D$ is a discrete subset of $X$, then $D$ itself is the only dense subset of $D$, and $\big\{\{x\}:x\in D\big\}$ is a relatively open cover of $D$ with no smaller subcover. Both of these inequalities are stronger than the inequality $s(X)\le nw(X)$, because $hd(X),hL(X)\le nw(X)$; I’ll leave that as an easy exercise.
In particular, $s(X)=\omega$ if $X$ is either hereditarily separable or hereditarily Lindelöf. There are hereditarily separable and hereditarily Lindelöf spaces that do not have countable net weight; perhaps the most familiar example is the Sorgenfrey line $\Bbb S$, the real line topologized by taking $$\big\{[a,b):a,b\in\Bbb R\text{ and }a<b\big\}$$ as a base for the topology. Thus $\Bbb S$, like $\Bbb R$, has no uncountable discrete subset, though $\Bbb S^2$, unlike $\Bbb R^2$, does have one, namely, $\big\{\langle x,-x\rangle:x\in\Bbb S\big\}$.
