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Here is a proof that does not use any transfinite induction at all!
Take any uncountable closed subset $S$ of $\mathbb{R}$.
Let $S_n = S \cap [n,n+1]$ for each integer $n$.
Then $S_n$ is uncountable for some integer $n$, so we can assume that $S \subseteq [a,b]$ for some reals $a,b$.
Let $p...