A set $S \subset \mathbb{R}$ has no cluster points iff $S \cap [-n,n]$ is a finite set for each $n \geq 1$
Possible useful definitions:
Defn: A set $A$ is compact $\Leftrightarrow$ the set is closed and bounded.
Defn: A set is closed if it contains all of its limit points
Defn: A point $x$ is a cluster point of a subset $A$ of $\mathbb{R}^n$ if there is a sequence$(a_n)_{n = 1}$ with $a_n \in A\setminus \{x\}$ such that $x = \lim_{n\to\infty}a_n$
Attempt: I don't really have an idea of how to formally prove this, but I will write out what I think would be the way to go about this.
$\Longrightarrow$ If I assume that the set $S$ has no cluster points, then that would mean that the set is not closed. This means it is not compact. Now I feel that from this there must be some way to create a one to one function with $\mathbb{N}$ that may possibly show finiteness, but I am struggling to do such.
$\Longleftarrow$ I am not really sure how to start. The question I am asking myself is, what would the set $S \cap [-n,n]$ look like? and if the set is finite how can I show it is not closed since I beleive not having cluster points would imply.
$\Longrightarrow$ By contraposition. Suppose that it exists $n \ge 1$ such that $S \cap [-n,n]$ is infinite. Therefore, it exists a sequence $(s_m)$ of distinct points all belonging to $S \cap [-n,n]$. As $[-n,n]$ is compact, $(s_m)$ has a limit point which is a cluster point of $S$.
$\Longleftarrow$ Again by contraposition. Suppose that $s$ is a cluster point of $S$. There is a sequence $(s_n)$ of points of $S \setminus \{x\}$ converging to $s$. The sequence is bounded, let’s say by $M>1$ integer. The sequence has an infinite number of points as it converges to a limit which is not in its range. As $s_n \in [-M,M]$ for all $n \in \mathbb N$, $S \cap [-M,M]$ is infinite.