Q:

example of a sequence $x_n$ such that the set of limit points of $x_n$ is $[0,1]$.

Give an example of a sequence $x_n$ such that the set of limit points of $x_n$ is $[0,1]$.
I know the answer is $\mathbb{Q}$ intersection $[0,1]$, but how can I prove that? 
Can you help me please? 

A:

The rationals in $[0,1]$ are countable, so they can be enumerated as a sequence $\{q_n\}$. Since the rationals in $[0,1]$ are dense in $[0,1]$, its limit point set is the whole $[0,1]$.
The only step you need to finish is the denseness property. Now observe that any point $x$ in $[0,1]$ has a decimal representation $x=0.a_1a_2\dots$. Then each $x_n=0.a_1a_2\dots a_n$ is rational, and $\lim_{n\to\infty}x_n=x$. So $x$ is a limit point of $\mathbb{Q}\cap [0,1]$.

