For any $F:{\mathcal P}(Y)\to Y$ there is a unique a unique well-ordering $(W, \lt)$ with $W\subseteq Y$ such that:
(Update: In general, the answer to the question is no. See here.)
Update, Sep. 6, 2017: Let me add a few additional remarks. First, in comments, Martin Brandenburg asked why one should bother about trying to obtain a "diagonalization-free" proof. That the proof above avoids diagonalization is perhaps simply a curiosity (though one is left with the question of how to define precisely "diagonalization-free"); what matters is that the argument gives a bit more than Cantor's: As I pointed out in a comment, the proof just given shows that 1) The collection of well-orderable subsets of $X$ has strictly larger size than $X$. This is an improvement over Cantor's result in the context of $\mathsf{𝖹𝖥}$. 2) Given any $f\!:\mathcal P(X)\to X$, we can find $A\subsetneq B$ with $f(A)=f(B)$. This is also a combinatorial strengthening, and it can be pushed further. Stevo Todorcevic in particular obtained several extensions of this idea, see this answer in Math.Stackexchange.
Second, Hartogs's theorem can be used to provide a different (also "diagonalization-free") proof of Cantor's result, and actually establish a generalization in the context of quasi-ordered sets, due to Gleason and Dilworth. For the pretty argument and appropriate references, see here.
