Every well-ordered set $W$ is isomorphic to a unique ordinal
Proof: We start with some well-ordered set $W$. We attempt to construct a bijective map from $W$. Let us consider this class $$\{(x, \alpha) : x \in W \land W(x) \cong \alpha \land \alpha \in Ord\}$$ We simply pair an element in $W$ with an ordinal that is isomorphic to initial segment of $W$ given $x$. This class is a function. Indeed, let ordinals $\alpha, \beta$ both isomorphic to $W(x)$. They must be isomorphic to each other. But, $$\alpha \cong \beta \land \alpha, \beta \in Ord \Rightarrow \alpha = \beta$$ This proof Isomorphic ordinals proves it. My first question:
Consider this lemma: No well-ordered sets are isomorphic to initial segment of itself. How is this used as alternative proof?
Now, since $F$ is a function, then $F(W)$ is a set since $W$ is a set. We want to show that: $dom(F) = W$. $dom(F) \subset W$ is trivial. But why $W \subset dom(F)$. Then, there exists a least element $x$ of $W - dom(F)$. So, any $y < x$ would be in $dom(F)$, which is absurd. The trivial case is $\emptyset$ which is by definition $F(\emptyset) = \emptyset$.
Now, we can show that $F(W) = \gamma$ for some ordinal $\gamma \in Ord$.
Since $F(W)$ is a set, it must not $Ord$ which is proper class. So, there exists least $\gamma \in Ord$ such that $\gamma \notin F(W)$. But how $F(W) = \gamma$. Again, $\gamma \subset F(W)$ is trivial. But how $F(W) \subset \gamma$.
Let $\delta \in F(W)$. It is clear that $\delta \in \gamma$ or $\gamma \in \delta$ or $\gamma = \delta$. Assume that $\gamma \in \delta$. Then, $$\gamma \in \delta \land \delta \in F(W) \Rightarrow \gamma \in F(W)$$ , which is absurd.
$F(W)$ need not be ordinal to make this work. Since $\delta \in F(W)$, we must have $\delta \cong W(x)$ for some $x \in W$. Now, let $h$ be the isomorphism. Due to $\gamma \in \delta$, it is valid to write $h(\gamma)$. We do have $\gamma \cong W(h(\gamma))$. So, $\gamma \in F(W)$.
Finally, it is clear why $F$ is injective. Indeed, let $F(x_1) = F(x_2)$. Then, the fact $$W(x_1) \cong F(x_1) = F(x_2) \cong W(x_2)$$ shows that $x_1 < x_2$ or $x_2 < x_1$ fails the above lemma. Secondly, this proof for order-preserving might work:
It is order-preserving because with $x < y$, we must have $W(x) \subset W(y)$. Let $h: W(y) \leftrightarrow F(y)$. $$x < y \Rightarrow W(x) \subset W(y) \Rightarrow h(W(x)) \subset h(W(y)) \Rightarrow h(W(x)) \subset F(y)$$
How to make $h(W(x)) = F(x)$? How to overcome these points to make the proof work? Am I following wrong approach?
