If $S\subseteq R$ is uncountable then $S$ has a super limit point

$x\in S$ is said to be a super limit point if $\forall \epsilon >0, (x-\epsilon,x+\epsilon)\cap S$ is uncountable.

I wondered if $S\subseteq R$ is uncountable then $S$ has a super limit point? I can prove this in the case $S$ is closed: One of the intervals $([n,n+1], n\in \mathbb{Z})$ contains uncountably many points of $S$, wlog $[0,1]$. Then one of the intervals $[0,{1\over2}] ,[{1\over2},1]$ contains uncountably many points of $S$, call it $I_1$. Now keep subdividing inside $I_1$ in the same way. $J_i=S\cap I_i$ is compact so the intersection is nonempty and in fact is contained in $S$ by closure. Then clearly any point in this intersection is a super limit point.

What happens when $S$ isn't closed?

Assume not, that S has no super limit points. Then for each $x\in S$ let $I_x$ be an open interval with rational endpoints such that $I_x\cap S$ is countable. Then $S=\cup (S\cap I_x)$ is countable, as there are only countably many distinct $I_x$.