 Solution

Suppose $f:S\to \mathcal{P}(S)$ were a surjective function. Form the subsetSince $D$ belongs to the power set, surjectivity would give $D=f(d)$ for some $d\in S$. But then $d\in D$ holds if and only if $d\notin f(d)=D$, a contradiction. Thus no such function is surjective, and in particular there is no bijection from $S$ to its power set. This is Cantor theorem; the argument also applies when $S$ is empty.

For infinite binary sequences, use the bijectionwith $\mathcal{P}(\mathbb{N})$, taking $\mathbb{N}=\{1,2,\dots \}$. Conversely, a subset determines its sequence of membership indicators. The power set is infinite, since it contains all singletons, and Cantor theorem rules out a bijection with $\mathbb{N}$. Hence it is an uncountable set, as is the set of infinite binary sequences. Equivalently, a proposed enumeration $(x_n^{(j)}{)}_{n\ge 1}$ is defeated by the sequence $y_j=1-x_j^{(j)}$, which differs from its $j$th listed sequence at position $j$. This is the Cantor diagonal argument.