Is there a sequence with an uncountable number of accumulation points?

Since the rational numbers are countable, we know there is a bijection $f:\mathbb N\to\mathbb Q$. Let $x_n=f(n)$. Then this is a sequence which contains every rational number, and its set of accumulation points is $\mathbb R$.

Terms of a sequence $$(\sin n)_{n\in\mathbb N}$$ are dense in $[-1,1]$ so each number in the interval is an accumulation point of the sequence.