Is there a sequence of real numbers such that the set of its subsequential limits is $[0,1]$? 
Is there a sequence of real numbers such that the set of its subsequential limits is $[0,1]$?

I've been considering the sequence:
$$1,~~\frac12,~~\frac22,~~\frac32,~~\frac13,~~\frac23,~~\frac33,~~\frac43,~\ldots~,~~\frac1n,~~\frac2n,~~\frac3n,~\ldots~,~~\frac{n+1}{n},~\ldots$$
Does it satisfy the condition?

For the question of the title: Consider $\mathbb{Q} \cap [0,1]$.
For the question of the body: Yes. For any $x \in [0,1]$ you can construct a subsequence as follows: Pick the first element as $1$. Now, given the $i$-th element constructed, for the next consider the $\frac{1}{i+1}$-neighbourhood, pick a number with denominator greater than the last one and slide the numerator until it reaches the neighbourhood (you can guarantee that this is possible by picking a large enough denominator by the Archimedean property). 
