Compactness and ordinals.

Solution 1:

As I mentioned in the comments, your attempted proof of the first part (that $[0, \Omega)$ is not compact) does not make any sense at all. You need to propose a specific open cover of $[0, \Omega)$ and then prove that this open cover does not admit a finite subcover.

For the second part, we actually don't need to know anything about $\Omega$ other than that it's an ordinal. You should consider an arbitrary open cover $U$ of $[0, \Omega]$. Let $S = \{x \in [0, \Omega] : $ for no $U_1, U_2, \ldots, U_n \in U$ is it the case that $[0, x] \subseteq U_1 \cup U_2 \cup \cdots \cup U_n\}$. Suppose that there is no finite subcover of $U$. Then $\Omega \in S$. Since $S$ is a nonempty set of ordinals, it must have a least element $\xi$. Now take some $K \in U$ such that $\xi \in K$. Then take some $(a, b) \subseteq U$ such that $a < \xi < b$. Then since $a < \xi$, we must have $a \notin S$, and therefore there is some $U_1, U_2, \ldots, U_n \in U$ such that $[0, a] \subseteq U_1 \cup \cdots \cup U_n$. But then $[0, \xi] \subseteq U_1 \cup \cdots \cup U_n \cup U$, which contradicts the claim that $\xi \in S$.