Proof

Let $S$ be a set and $\mathcal P(S)$ its power set.
If $S=\emptyset$ is empty then $\mathcal P(S)=\{\{\emptyset\}\}.$ There is no function $f:S\to\mathcal P(S)$, especially no surjective function.
Thus, assume $S$ is not empty.
In this case we can define some function $f:S\to\mathcal P(S)$, i.e. for every $s\in S$ we have $f(s)\in\mathcal P(S).$
Suppose, $f$ is surjective.
We construct the set $X=\{s\in S\mid s\not\in f(s)\}$.
Since $f$ is surjective and $\mathcal P(S)$ contains the empty set $\emptyset$ as its element, there is by definition of surjectivity at least one $s_1\in S$ with $f(s_1)=\emptyset.$ Since $s_1\not\in\emptyset$, we have $s_1\not\in f(s_1)$ and therefore $X$ is not empty, because it contains at least $s_1.$
Note that $X$ is also a subset of $S$, thus $X\in\mathcal P(S).$ Again, since $f$ is surjective, there is at least one $s_2\in S$ with $f(s_2)=X.$
Now, there are two cases:
- $s_2\in X$ - this case is not possible, since by definition of $X$ we have $s_2\not\in f(s_2)=X.$
- Therefore, we must have $s_2\not\in f(s_2)=X.$ But either in this case it follows from the definition of $X$ that $s_2\in X.$
The assumption has to be wrong, i.e. $f$ is not surjective.
∎

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Knauer Ulrich: "Diskrete Strukturen - kurz gefasst", Spektrum Akademischer Verlag, 2001
Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994