Showing that $[0, \omega_1]$ is compact. I want to show that $[0, \omega_1]$  is compact, where $\omega_1$ is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals. 
The tips I have seen to showing this is that $[0, \omega_1]$:


*

*contains a maximal element

*there are no infinite strictly decreasing sequences of ordinals.  


Take an open cover of the desired set. My thoughts are that some open set must contain $\omega_1$, which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
 A: We can show for an arbitrary ordinal $\alpha$ that $[0,\alpha]$ is compact with the order topology.
Let $\mathcal U$ be an open cover of $[0,\alpha]$, and let $\beta$ be the smallest ordinal such that $[\beta,\alpha]$ is covered by a finite subcover of $\mathcal U$. (Certainly there are some $\beta$s with this property, such as $\alpha$ itself, so there must be a smallest one).
Now if $\beta$ is a successor ordinal $\beta=\delta+1$, then there must be an element of $\mathcal U$ that contains $\delta$, and adding that to our finite cover of $[\beta,\alpha]$ produces a finite cover of $[\delta,\alpha]$, a contradiction.
On the other hand, if $\beta$ is a limit ordinal, then every open set that contains $\beta$ must also contain some $\delta<\beta$, and therefore our finite subcover of $[\beta,\alpha]$ is already a cover of $[\delta,\alpha]$. Again a contradiction.
The only possibility remaining is that $\beta=0$.
A: Suppose that $\mathcal{U}$ is an open cover of $[0,\omega_1]$ by basic open sets of the order topology (that's the topology we are putting on this ordered set), and suppose for a contradiction that it has no finite subcover. (It suffices to consider bcovers by base elements, as is easy to see in all topological spaces, if you know Alexander's subbase theorem, the proof will be easier even).
Some element of the cover has to contain the maximale element $\omega_1$, and so some $U_0 \in \mathcal{U}$ must be of the form $(\beta_0, \omega_1]$ (such are the basic open subsets in the order topology that contain the maximal element).
Also, $0$ must be covered so there is some $U'_0 \in \mathcal{U}$ of the form $[0,\alpha_0)$ that contains $0$. 
Now we know that $U_0, U'_0$ cannot be a finite subcover (which would happen when $\alpha_0 > \beta_0$), and so that some $U_1 \in \mathcal{U}$ must cover $\beta_0$, say $\beta_0 \in U_1 = (\alpha_2, \beta_1)$, which implies that $\alpha_2 < \beta_0 < \beta_1$.
Now consider what happens if the sets $U_0, U'_0, U_1$ do not form a subcover yet, find a new element to cover etc., and define an infinite  decreasing sequence of ordinals that way.
