Proving Cantor's theorem $\DeclareMathOperator{\card}{card}$From Problem-Solvers Topology

Prove the following:
CANTOR'S THEOREM
If $A$ is a set, then $$\card A < \card \mathcal{P}(A)$$
where $\card A$ stands for the cardinality of set $A$.

My Answer
If $A = \emptyset$, then $\card A =0$ and $\card \mathcal{P}(A)=1$.
If $A = \{a\}$, then $\card A=1$ and $\card \mathcal{P}(A)=2$.
So suppose that $A$ has at least two elements.
Define a function $f: A \rightarrow \mathcal{P}(A)$ such that $f(x) = \{x\}$ for all $x \in A$.
Then $f$ is injective. But it cannot be surjective, because for any two distinct elements $a,b \in A$, there is no element in $A$ that is sent to the set $\{a,b\}$ in $\mathcal{P}(A)$. Therefore, there is no bijection, and $\card A < \card\mathcal{P}(A)$.
Do you think my answer is correct?
Thanks in advance

$\DeclareMathOperator{\card}{card}$No. Your answer is not correct. Between any two sets (both having two elements or more) there is a non-surjective map. Your task is to show that every set $A$ and every function from $A$ to $\mathcal P(A)$ is not surjective.
To clarify the point you're missing, the argument in your proof amounts to the following "proof":

"Theorem": There is no bijection between $\{0,1\}$ and $\{2,3\}$.
"Proof". Consider the function $f(0)=f(1)=2$. It is clearly not surjective, therefore it is not a bijection. And therefore $\card(\{0,1\})\neq\card(\{2,3\})$.

