Cantor's Power Set Theorem; uncountable sets

Cantor's Power Set Theorem says that if $\mathbb{S}$ is any set, then there is an injection from $\mathbb{S}$ to $\mathcal{P}(\mathbb{S})$ but no bijection, so $\vert\mathbb{S}\vert< \vert\mathcal{P}(\mathbb{S})\vert$.
In particular it follows that $\mathcal{P}(\mathbb{N})$ is uncountable.

The first part of the theorem is easy:

• For any set $\mathbb{S}$, the map $p: s \rightarrow p(s) = \{s\}$, defined for any $s \in \mathbb{S}$ gives a well-defined injection from $\mathbb{S} \rightarrow \mathcal{P}(\mathbb{S})$.

The second part goes as follows.

• Let $q: \mathbb{S} \rightarrow \mathcal{P}(\mathbb{S})$ be a map.
  Put $\mathbb{Y} = \{ x \in \mathbb{S}: x \notin q(x)\}$.
  Then $\mathbb{Y}$ is a well-defined subset of $\mathbb{S}$.
  We prove that no $y \in \mathbb{S}$ exists, such that $q(y) = \mathbb{Y}$, for some $y \in \mathbb{S}$.
  Suppose the contrary so there is $y \in \mathbb{S}$ with $q(y) = \mathbb{Y}$.
  We ask: is $y \in q(y) = \mathbb{Y}$?
  • If $y \in q(y) = \mathbb{Y}$, then $y \notin q(y)$ is false, so by definition of $\mathbb{Y}$ we have $y \notin \mathbb{Y}$, a contradiction.
  • If $y \notin q(y) = \mathbb{Y}$, then $y \notin q(y)$ is true, so by definition of $\mathbb{Y}$ we have $y \in \mathbb{Y}$, a contradiction.
  So both the conditions $y \in \mathbb{Y}$ and $y \notin \mathbb{Y}$ lead to a contradiction.
  So the assumption that $q(y) = \mathbb{Y}$ for $y \in \mathbb{S}$ leads to a contradiction.
  So no such $y$ exists.
  So the set $\mathbb{Y}$ is not in the range of $q$.
  So $q$ is not surjective.

We have proved that for any set $\mathbb{S}$, there is no surjection from $\mathbb{S} \rightarrow \mathcal{P}(\mathbb{S})$, so no bijection either and we are done.