Need a hint: show that a subset $E \subset \mathbb{R}$ with no limit points is at most countable.

Hint. $\mathbb{Q}$ is dense in $\mathbb{R}$. Say there is an uncountable set without limit points. How is $\mathbb{Q}$ shared in their neighborhoods?

For $n\geq1$ let $E_n:=E\cap[{-n},n]$. As $E=\bigcup_{n\geq1} E_n$, it is enough to show that each $E_n$ is countable, or even finite. Let an $n\geq1$ be given, and consider the set $I:=[{-n},n]$. Each $x\in I$ possesses an open neighborhood $U(x)$ that contains at most one point of $E$. The family $\bigl(U(x)\bigr)_{x\in I}$ is an open covering of $I$, and as $I$ is compact there are $N$ points $x_k\in I$ such that $I\subset\bigcup_{k=1}^N U(x_k)$. Since each $U(x_k)$ contains at most one point of $E$ it follows that $$\#E_n=\#(E\cap I)\leq\sum_{k=1}^N E\cap U(x_k)\leq N\ ,$$ as claimed.

proof by contradiction. suppose E is uncountable. consider the set E intersection with [n,n+1] for integers n. then there exist at least one n such that the above type of set is uncountable. clearly that set is bounded in real nos. then by Bolzano Weierstrass's theorem the above set has a limit point in R. a contradiction to the hypothesis.

Note: The union of countably many countable sets is countable.

Assume $E$ is not countable. Then one of the sets $[n,n+1]\cap E$ with $n\in \mathbb Z$ is uncountable. Starting with $a_0=n$, $b_0=n+1$ we find a sequence of nested intervals $[a_k, b_k]$ such that $[a_k,b_k]\cap E$ is uncountable. In fact we can simply bisect an interval at each step and note that at least one of the halves must have iuncountably many points in common with $E$. In other words, we let $a_{k+1}=a_k$, $b_{k+1}=\frac{a_k+b_k}2$ if $[a_k, \frac{a_k+b_k}2]$ is uncountable and let $a_{k+1}=\frac{a_k+b_k}2$, $b_{k+1}=b_k$ otherwise (and observe that then $[a_{k+1},b_{k+1}]\cap E$ is uncountable as well. The nested intervals contain a point $c\in \mathbb R$. This $c$ has some (in fact uncountably many) points of $E$ in every $\epsilon$-neighbourhood. Indeed $[a_k,b_k]$ is contained in the $\epsilon$-neighbourhood as soon as $2^{-k}<\epsilon$.

Or: If $E\cap [0,\infty)$ and $E\cap(-\infty,0]$ are both countable, then so is $E$. Hence assume wlog that $E\cap [0,\infty)$ is uncountable. Let $a=\inf\{x\in \mathbb R\colon [0,x]\cap E\mathrm{\ is\ uncountable}\}$ If $a=\infty$ then $E\cap [0,\infty)=\bigcup_n E\cap[0,n)$ is the union of countable sets, hence countable. If on the other hand $a$ is finite, then $[a,a+\epsilon)\cap E$ is uncountable for any $\epsilon>0$, hence $a$ is a limit point.

It’s enough to assume that $E$ has no limit points in $E$. Then for each $x\in E$ there is an $n(x)\in\Bbb N$ such that $(x-2^{-n(x)},x+2^{-n(x)})\cap E=\{x\}$; call this interval $U_x$. The rationals are dense in $\Bbb R$, so for each $x\in E$ we may choose a rational $q_x\in U_x$. Suppose that $E$ is uncountable. $\Bbb Q$ is countable, so there must be some $p\in\Bbb Q$ such that $E_0=\{x\in E:q_x=p\}$ is uncountable. And $\Bbb N$ is also countable, so there must be some $m\in\Bbb N$ such that $E_1=\{x\in E_0:n(x)=m\}$ is uncountable. But then $p\in U_x=(x-2^{-m},x+2^{-m})$ for each $x\in E_1$; equivalently, $|x-p|<2^{-m}$ for each $x\in E_1$.

Suppose that $x,y\in E_1$ with $p\le x<y$; then $x\in E\cap[p,y]\subseteq E\cap U_y$, contradicting the choice of $U_y$. A similar contradiction arises if there are $x,y\in E_1$ such that $x<y\le p$. Thus, $|E_1\cap[p,\to)|\le 1$ and $|E_1\cap(\leftarrow,p]|\le 1$, so $|E|\le 2$, contradicting the uncountability of $E_1$ and showing that $E$ must in fact be countable.