First, it suffices to show that the set of all countable sequences is not countable. That avoids having to deal with sequences where $x_n$ isn't defined for some $n$. Then we can define a very natural map from that set to $[0,1)$ via $$ f \,:\, \{0,1\}^\mathbb{N} \to [0,1) \,:\, (x_n) \mapsto \sum_{k=1}^\infty x_n2^{-n} $$
If you can show that $f$ is surjective, and already know that $[0,1)$ has the same cardinality as $\mathbb{R}$, then it follows that $\{0,1\}^\mathbb{N}$ also has at least that cardinality.