Theorem 1   Every bounded, infinite set of real numbers has a limit point.

Proof. Let $S\subset\mathbb{R}$ be bounded and infinite. Since $S$ is bounded there exist $a,b\in\mathbb{R}$ with $a<b$ such that $S\subset[a,b]$ Let $b-a=l$ and denote the midpoint of the interval $[a,b]$ by $m$ Note that at least one of $[a,m],[m,b]$ must contain infinitely many points of $S$ select an interval satisfying this condition, denoting its left endpoint by $a_1$ and its right endpoint by $b_1$ Continuing this process inductively, for each $n\in\mathbb{N}$ we have an interval $[a_n,b_n]$ satisfying \begin{equation} [a_n,b_n]\subset[a_{n-1},b_{n-1}]\subset\cdots\subset[a_1,b_1]\subset[a,b]\text{,} \end{equation}where, for each $i\in\mathbb{N}$ such that $1\leq i\leq n$ the interval $[a_i,b_i]$ contains infinitely many points of $S$ and is of length $l/2^i$ Next we note that the set $A=\{a_1,a_2\ldots,a_n\}$ is contained in $[a,b]$ hence is bounded, and as such, has a supremum which we denote by $x$ Now, given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $x-\epsilon<a_N\leq x$ Furthermore, for every $m\geq N$ we have $x-\epsilon<a_N\leq a_m\leq x$ In particular, if we select $m\geq N$ such that $l/2^m<\epsilon$ then we have \begin{equation} x-\epsilon<a_n\leq a_m\leq x\leq b_m=a_m+\dfrac{l}{2^m}<x+\epsilon\text{.} \end{equation}Since $[a_m,b_m]\subset(x-\epsilon,x+\epsilon)$ contains infinitely many points of $S$ we may conclude that $x$ is a limit point of $S$