In Herrlich on page 5 he gives a proof of $\textbf{AC} \implies \textbf{WOT}$:
He does not give a definition of cardinality $|X|$ before this proof and I searched the index for a definition but couldn't find one. Hence, since we are in $\textbf{ZF}$ without $\textbf{C}$ I assume he uses the definition: $\alpha = \min \{\beta \mid \exists s \in V_\beta \text{ s.t. } s \text{ is in bijection with } X \}$ and $|X| = \{ s \in V_\alpha \mid s \text{ is in bijection with } X \} $.
My question then is the following: Why does one resort to Hartogs number for the proof? Can one prove it as follows: Let $\alpha$ as above be the rank of $X$. Then there cannot be an injection from $V_\alpha$ into $X$. Now replace $\aleph$ with $V_\alpha$ in the proof above. Et voilà, we shortened the proof by one definition. What am I missing? I am as always very grateful for your help. Thank you in advance.
Using $V_\alpha$ is not good because we don't know whether or not $V_\alpha$ can be well-ordered. In fact, if $X$ cannot be well-ordered and $X\in V_\alpha$ then it is impossible that $V_\alpha$ can be well-ordered.
What Herrlich is doing here is to define a surjection from $\aleph(X)$ onto $X\cup\{\infty\}$ which has the property that the only point in the range which has more than one point in its preimage is $\infty$, so $X$ has a bijection with an ordinal.
Doing the same thing with $V_\alpha$ to begin with will not allow us to conclude that there is an ordinal whose range is exactly $X$, which is how we prove that $X$ can be well-ordered.