Does my attempt look fine or contain gaps? Thank you so much!
Theorem: Every set of ordinals is well-ordered with respect to $\in$.
Lemma 1: Every nonempty set of ordinals has a least element with respect to $\in$.
Proof:
Let $X$ be a nonempty set of ordinals. Take $\alpha\in X$.
- $\alpha \cap X=\emptyset$
Suppose that $\beta \in X$ and $\beta\neq\alpha$. I claim that $\alpha\in\beta$. If not, $\beta\in\alpha$. Moreover, $\beta\in X$. Then $\beta\in\alpha \cap X$. This is a contradiction.
Hence $\forall (\beta\in X \text{ and }\beta\neq\alpha):\alpha\in\beta$. Thus $\alpha$ is the least element of $X$.
- $\alpha \cap X\neq\emptyset$
Then $\alpha \cap X$ is a nonempty subset of $\alpha$ and thus has a least element since $\alpha$ is well-ordered.
Let $\delta$ be the least element of $\alpha \cap X$. For all $\theta\in X$:
- $\theta\in \alpha$
Then $\theta\in \alpha \cap X$ and thus $\delta\in\theta$ since $\delta$ is the least element of $\alpha \cap X$.
- $\theta= \alpha$
Then $\delta\in\alpha \cap X=\theta\cap X$ and thus $\delta\in\theta$.
- $\alpha\in\theta$
Then $\delta\in\alpha\in\theta$ and thus $\delta\in\theta$.
Hence $\delta$ is the least element of $X$.
Lemma 2: For any two ordinals $\alpha$ and $\beta$. Either $\alpha\in\beta$, or $\alpha=\beta$, or $\beta\in\alpha$.
Proof:
I presented a proof here.
We proceed to prove our main theorem.
Let $X$ be a nonempty set of ordinals. Then $X$ is linearly ordered by Lemma 2. Every nonempty subset of $X$ has a least element by Lemma 1. Hence $X$ is well-ordered.
Your proof is fine. If you know that every subset of a set well-ordered by $\in$, is itself well-ordered by $\in$ then there is a fast way to prove your claim: Let $\beta =\bigcup_{\alpha\in X}(\alpha+1)$. Then $\beta $ is an ordinal and $X\subseteq \beta,\ $ which implies that $X$ is well-ordered by $\in$ since $\beta$ is.