If there is an injection $A\to B$, we say that $|A|\leq |B|$.
Proposition. If there exists a surjection $f\colon X\to Y$, then $|Y|\leq |X|$.
The proof goes as follows
Let $c$ be a choice function for $\mathscr{P}(X)$; since $\{f^{-1}(y)\mid y\in Y\}$ is a partition of $X$, it follows that $y\mapsto c\big(f^{-1}(y)\big)$ is an injection $Y\to X$.
My problem is that I don't know what the domain for $c$ is. A choice function was defined as an object of the form $c\colon \mathscr{A}\to \bigcup\{A_\alpha\mid \alpha\in\mathscr{A}\}$ that satisfies $\forall \alpha\in\mathscr{A},\; c(\alpha)\in A_\alpha$. The indexing set ($\mathscr{A}$) is not specified on the proof. What is the domain of $c$? Clearly $f^{-1}(y)$ must be an element of the indexing set. How was the indexation done such that $c(f^{-1}(y))=c(f^{-1}(\gamma))\implies y=\gamma$? This doesn't seem trivial since all we know is that $c(f^{-1}(y))\in A_{f^{-1}(y)}$ (this doesn't say anything about the set $A_{f^{-1}(y)}\subseteq \mathscr{P}(X)$).
Explicity, the choice function is $c:\mathcal{P}(X)\to\bigcup\{X: X\in \mathcal{P}(X)\}$, i.e. it picks an element from any subset of $X$.
Injectivity of $c \circ f^{-1}$ follows from the fact that $f^{-1}(y)$ and $f^{-1}(z)$ are disjoint sets whenever $y\neq z$. Thus, $c$ can't pick out a common element from the two sets.