Regarding the cardinality of all functions $\mathbb{R} \rightarrow \{0;1\}$, one can prove that the corresponding set $C$'s cardinality is strictly greater than that of $\mathbb{R}$ by assuming there is an injective map $f: C \rightarrow \mathbb{R}$, and looking at the map $g: x \mapsto 1 - f^{-1}(x)(x)$, where $f^{-1} \circ f = Id_C$.

$g \in C$ so $g(f(g))$ makes sense, but $g(f(g)) = 1 - f^{-1}(f(g))(f(g)) = 1 - f \circ f^{-1}(g)(f(g)) = 1 - g(f(g)) \neq g(f(g))$, which is impossible. So there can't be an injective map from $C$ to $\mathbb{R}$.

This can be generalized to any set instead of $\mathbb{R}$, though you would have to treat the case of the empty set separately.