Is there a sequence whose set of subsequential limits is $[0,1]$? 
Is there a sequence whose set of subsequential limits is $[0,1]$?

My guess is yes: I tried $s_n = |\cos n|$. It makes since intuitively that all the subsequences hit every number in $[0, 1]$ but I am having trouble proving this rigorously. 
 A: For this problem, I will suggest tackling a different problem first:
What are the properties of the set of limit points?
Hint: It's a closed set.
This makes the problem a little bit easier, all you need to do is find a sequence whose limit points form a dense set in [0,1].
Building such sequence is quite intuitive: Take for instance the sequence $a_1 = 1 , a_2 = 1/2$
$ a_3 = 1, a_4 = 2/3, a_5 = 1/2, a_6 = 1/3$

$ a_7 =1 \dots$
Edit: Btw: there is a beautiful theorem (I think by d'Alembert) stating:

THM: Let $\alpha \in \mathbb R \setminus \mathbb Q$. Then the set $ \{n\alpha -[n\alpha ]\colon n\in \mathbb N\}$ is dense in $[0,1]$.

$n\alpha -[n\alpha ] = \{n\alpha\}$ is the fractional part of $n\alpha$.
A: $s_1=0/1$, 
$s_2=0/2, s_3=1/2,$
$s_4=0/4, s_5=1/4, s_6=2/4, s_7=3/4$
$s_8=0/8, s_9=1/8, s_{10}=2/8, s_{11}=3/8, s_{12}=4/8, s_{13}=5/8, s_{14}=6/8, s_{15}=7/8,...$
 so on and so forth. 
Notice that $s_{2^n}=0$ and $s_{2^n+k}=k/2^n$ until $k$ gets as big as $2^n$, which then that equals to $s_{2^{n+1}}=0$
So this allows you to explicitly write out $s_n$ (not necessary for the proof)
take any $L \in [0,1]$. $\cfrac{a_n}{2^n}\leq L<\cfrac{a_n+1}{2^n}$ for any $n$, $a_n$ being some integer  
pick the subsequence $\bigg(\cfrac{a_n}{2^n}\bigg)$, this subsequence converges to $L$ 
