Prove that if $S$ is a finite set then $S$ has no limit points.

(You really should specify that $S\subseteq\Bbb R$; the statement isn’t true in topological spaces in general.)

Your argument is incomplete: it shows that no point of $S$ can be a limit point of $S$, but it doesn’t show that $S$ has no limit points in $\Bbb R$. Suppose, for instance, that $S=\{0,2\}$ and $x_0=1$. Then your $\epsilon=2$, and it’s not true that there is no $x\in S$ such that $|x-x_0|<2$; in fact, $|x-x_0|<2$ for both $x\in S$.

Your argument is fine if $x_0\in S$. If $x_0\notin S$, let $\epsilon=\min\{|x_0-a|:a\in S\}$. The distances $|x_0-a|$ are all positive, and there are only finitely many of them, so $\epsilon>0$, and it’s clear that there is no $x\in S$ such that $|x-x_0|<\epsilon$.

Re-check your definition of a limit point. A limit point for $S$ does not necessarily have to be in $S$. That should lead you to the right answer.

If $S=\{a_1,a_2\}\subseteq\mathbb{R}$, and $x$ lies halfway between $a_1$ and $a_2$, then your $\epsilon$ will not work.

Here is what works: Let $\epsilon$ be the smallest number in $\{|x-a_i|:i=1,\ldots,n,\,a_i\neq x\}$.