Prove that the size of any nonempty set $A$ is smaller than its powerset $P(A)$, i.e. $|A| < |P(A)|$. My approach to this problem is as follows.
First I consider all sets whose size is less than two, $|A| < 2$. Let $|A| = 1$, i.e. the case where $A$ has only one element $a \in A$. Then, the powerset of $A$, $P(A)$, contains two elements, the empty set and the single element $a$, i.e. $\{ \emptyset, a \}$. In this case $|A| = 1 < 2 = P(A)$ and the claim holds.
Now I consider all sets whose size is at least two, $|A| \geq 2$. Let $|A| \geq 2$. According to the definition of the powerset $P(A)$ of $A$, the powerset contains all elements of $A$, so $|A| \leq |P(A)|$. I search for elements in $P(A)$ not in $A$. I observe that the set $A$ is itself a subset of $A$ and so is an element of $P(A)$. This element is not in $A$ and so $|A| < |P(A)|$. So I conclude that for any set whose size is at least two I find an element in the powerset which is not present in $A$. And so we conclude that the size of any set $A$ is smaller than the size of its powerset. And so the claim holds in general.
The proof can be simplified by noting that for any set $A$ the empty set $\emptyset$ is a member of the powerset $P(A)$, but not of $A$, i.e. $\emptyset \notin A$, and since each element in $A$ is also in the powerset the size of the powerset $P(A)$ is at least one greater than the size of $A$, i.e. $|A| + 1 \leq |P(A)|$ which imples $|A| < P(A)$, and so the claim holds.
Now it should be stated that the proof relies on the assumption that $A$ contains no other sets as elements, e.g. the empty set.
So to summarize I use the fact that all elements in $A$ are also in $P(A)$ and $P(A)$ contains elements not in $A$ so its size is strictly greater than $A$.
Are there any flaws in the approaches I followed? And what could possible alternative approaches be?
Claim: Let $f: A \to \mathscr P(A)$ be a map. Then $f$ is not surjective. Note that the claim proves that $\vert A \vert \lt \vert \mathscr P(A) \vert$ even for infinite $A$.
Proof: Define $B=\{ x \in A \mid x \notin f(x) \} \in \mathscr P(A)$. Claim: $B \notin f(A)$. If, contrary to the Claim, $B=f(y)$ for some $y \in A$, then $y \in B \Rightarrow y \notin f(y)=B$ by the definition of $B$. That contradiction shows that $y \in f(y)$ is not possible. On the other hand, $y \notin f(y)\Rightarrow y \in B=f(y)$, again by the definition of $B$. That contradiction shows $y \notin f(y)$ is not possible.
If $y$ exists, however, it must be the case that either $y \in B$ or $y \notin B$, and we have now shown that neither of these alternatives is possible. Our initial assumption was that $\exists y~(f(y)=B)$. This assumption must therefore be false, so $f$ cannot be surjective.