Cantor's theorem

In the proof that $|S| < |\mathcal P({S})|$, the set $\{x \in S|x \notin f(x)\}$, where $f: S \rightarrow \mathcal P(S)$ is used.

I understand the method and all, but how can you ensure such a set exists in the first place? During the construction of such a function can't you make sure that $x \mapsto x \in f(x) \subseteq \mathcal P(S)$ and thus ensure that $\{x \in S|x \notin f(x)\} = \emptyset$? If you were to say a set definitely must exist, then aren't you straightout claiming that the cardinality of $\mathcal P(S)$ must be greater?

Sorry, I'm just really confused about this at the moment.

The empty set exists, and being the empty set does not cause a problem.

You can ensure it is empty. It doesn't matter.

Firstly, the set exists because it is well defined (so long as $S$ and $f$ are well defined).

Secondly, here's an example to show it doesn't matter if that set is empty. Let's look at the set $S=\{1\}$. Then $\mathcal{P}(S) = \{\emptyset, \{1\}\}$

Take $f: S\mapsto \mathcal{P}(S), 1 \mapsto \{1\}$. Then now of course $\{x\in S : x\not\in f(x) \} = \emptyset$. But what in S is left to map to the empty set!? If the situation were different, and we mapped 1 to the empty set instead; then $\{1\}$ would have no preimage under f.

The clinch of the proof is that for this set, if it somehow did have a preimage, then the elements in its preimage will cause contradictions about memeberships. That's what makes the proof so clever. Cantor spotted a well defined subset of a set under any map to the power set, which can't have a preimage.