elementary set theory - Question about cardinality of some set of functions

The question in its original form deals with the problem of deciding whether the set $T$ of all irrational numbers in the set $[0,1]$ such that they have only digits $0$ and $1$ in their decimal expansion is countable or uncountable?

I have been thinking along this lines and this form of the problem deals with the question in the title, if we look at the set $S$ of all functions $f: \mathbb N \to \{ 0,1 \}$ then this set of functions describe all numbers in $[0,1]$ which have $0$ and $1$ in their decimal expansion, and if $S$ is countable then obviously $T$ is also countable as a subset of $S$ but if $S$ is uncountable then $T$ is uncountable because $R=S\setminus T$ is the set of all rational numbers in the $[0,1]$ that have $0$ and $1$ in their decimal expansion so obviously $R$ is countable as a subset of $\mathbb Q$, which is countable.

So how to decide whether $S$ is countable or uncountable?

Answer

The set $S$ is often denoted as $2^\Bbb N$, and it is not hard to show that $2^\Bbb N$ has the same cardinality as $\mathcal P(\Bbb N)$, by mapping a subset to its indicator function.

Use Cantor's theorem to conclude that $S$ is uncountable.

One can actually show further that $|[0,1]|=|\mathcal P(\Bbb N)|$, and that in fact $T$ has the same cardinality as $[0,1]$ as well.