Question:
Let $X$ be a topological space in which every open set is compact. Show that:
a. Every ascending chain of open sets in $X$ stabilizes: If $U_1\subseteq U_2\subseteq U_3\subseteq\dots$, $\exists n\in \mathrm{N} : U_n=U_m, \forall m>n$.
b. If $X$ is Hausdorff, then it has a discrete topology. (Recall that compact sets in a Hausdorff space are closed)
What I tried:
For the first item, consider $\cup_{i=1}^{\infty} U_i$. It's open, thus compact. If the chain didn't stabilize, there would be no finite subcover, contradicting its compactness.
For the second item, I know that I must show that every subset of $X$ is open. I haven't studied irreducibility yet, but this seems close to what I need to show. I understand that a finite set that is Hausdorff has the discrete topology, so I wonder if there is way to not use the irreducibility related stuff in the answer in the link above and show $X$ has to be finite.
If, as in the first item, we consider $\cup_{i=1}^{\infty} U_i$, we can write it in terms of disjoint open sets: $\cup_{i=1}^{n} U_i= \cup_{i=1}^{n} U_i - \cup_{j=0}^{i-1} U_j$. But I'm not sure if this is useful.