Infinitely many accumulation points in a bounded sequence?

1, 1,1/2, 1,1/2,1/3, 1,1/2,1/3,1/4, ...

Yes. Take the sequence that enumerate all rational between $[-1,1]$.

A possible sequence that enumerate rational in $[0,1]$: $$x_{\frac{q(q-1)}{2}+p}=\frac{p}{q}.$$

Take an infinite family of bounded sequences in $[0, 1]$ converging to $1/k$ for each $k \in \mathbb{N}$.

Interweave the sequences (dovetail them) to produce an explicit example.

That is, for $a_k \rightarrow 1$, $b_k \rightarrow 1/2$, $c_k \rightarrow 1/3$, $\ldots$:

$a_1, a_2, a_3, a_4 \ldots$

$b_1, b_2, b_3, b_4 \ldots$

$c_1, c_2, c_3, c_4, \ldots$

$\vdots$

becomes:

$a_1, b_1, a_2, c_1, b_2, a_3, \ldots$

where this final sequence has the desired property.