Let $\Omega$ denote the minimal uncountable well-ordered set. We are trying to prove that every sequence $\{x_n\} \subset \Omega$ is bounded above. To begin, assume to the contrary that such an unbounded sequence $(x_n)$ is indeed fully contained in $\Omega$. $S_{x_n}$, the section of $x_n$, is countable for any $n$ by definition of $\Omega$. Then the union of all such $(x_n)$ is countable, as it is just a countable union of countable sets. To prove a contradiction, we show that $\cup S_{x_n} = \Omega$. Forward set inclusion is trivial, as $S_{x_n} \subset \Omega$ for all $n$ by assumption. In the reverse direction, for any $y \in \Omega$, there exists a $N$ such that $y \in S_{x_N}$ since $(x_n)$ is not bounded above. Therefore, $\Omega \subset \cup S_{x_n}$ and we've shown the sets are equal. However, $\Omega$ is uncountable by definition while $\cup S_{x_n}$ is countable, so we've a contradiction.
Is this proof correct?