Sequence whose limit points are all the points of $[0,1]$ (CSIR)

Prove that there exists a sequence of real numbers whose limits points are points of $[0,1]$.
Proof :
Let $\{x_n\}$ be a sequence of real numbers and a real number $\alpha$ is said to be a limit point of $\{x_n\}$ if there exists a subsequence $\{y_n\}$ of $x_n$ such that $\{y_n\}$ converges to $\alpha$.
To answer the given problem, we need to find a sequence $\{x_n\}$ satisfying "given a point $\alpha$ in $[0,1]$ the sequence $\{x_n\}$ should have a subsequence $\{y_n\}$ that converges to $\alpha$". We shall construct one such sequence.

Fact : Rationals in $[0,1]$ form a countable dense subset of $[0,1]$.

Since $[0,1] \cap \Bbb Q$ is countable, we can enumerate them and write it as a sequence say $\{x_n\}$. Let $\alpha \in [0,1]$, since $[0,1] \cap \Bbb Q$ is dense in $[0,1]$ every point of $[0,1]$ is a limit point of $[0,1] \cap \Bbb Q$. Hence there is a sequnce of points of $[0,1] \cap \Bbb Q$ which converges to $\alpha$. This is clearly a subsequence of $\{x_n\}$ and the proof is complete.
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