I am reading "Set Theory and General Topology" by Fuichi Uchida.
In this book there is the following theorem:
Let $X$ be any set. There exists an order $\rho$ such that $(X, \rho)$ is a well-orderd set.
Proof:
Let $X$ be a set.
Let $\mathcal{W} := \{(A, \alpha) \mid A \subset X, \alpha \text{ is a well-order on }A\}$.
For $(A, \alpha), (B, \beta) \in \mathcal{W}$, we define $(A, \alpha) \leq (B, \beta)$ if and only if $(A, \alpha) = (B, \beta)$ or $(A, \alpha)$ is a section of $(B, \beta)$.
Then, $(\mathcal{W}, \leq)$ is a partially ordered set.
Let $\mathcal{W}^{'}$ be a totally ordered subset of $(\mathcal{W}, \leq)$.
Let $W := \bigcup \{A \mid (A, \alpha) \in \mathcal{W}^{'}\}$.
Then there exists a unique well-order $w$ on $W$ such that each $(A, \alpha) \in \mathcal{W}^{'}$ is equal to $(W, w)$ or is a section of $(W, w)$.
So, $(\mathcal{W}, \leq) $ is inductive.
$\cdots$
How can we proof that there exists a unique well-order $w$ on $W$ such that each $(A, \alpha) \in \mathcal{W}^{'}$ is equal to $(W, w)$ or is a section of $(W, w)$?
Then applying Zorn's Lemma to $\mathcal W$, we obtain a maximal element $(A,\alpha)$, which has to be a well-ordering in $X$.
Otherwise, if $(A,\alpha)$ is a maximal element and $x\in X\setminus A$, then setting $A'=A\cup\{x\}$ and $\alpha'$ extension of $\alpha$, with $x$ larger than all the elements of $A$, we obtain that $(A',\alpha')$ is another well-ordering of a subset of $X$ and $(A,\alpha)<(A',\alpha')$. Contradiction to the maximality of $(A,\alpha)$.