Proposition: Show that every singleton in a hereditarily compact Hausdorff space is clopen
Recall, a space is hereditarily compact if every subset is compact.
My attempt:
Let $x \in X$, then $\{x\}$ as a singleton is compact since every single subset is compact. It is compact in a Hausdorff space, so it is closed. Then given $x\neq y \in X$, $X \backslash \bigcup_{x \in X\backslash \{y\}} \{x\} = \{y\}$
The next step is that I wish to show $\{y\}$ is open. But this is only true if $X$ is finite $\implies \bigcup_{x \in X\backslash \{y\}} \{x\}$ is closed (finite union of closed is closed).
But to prove that $X$ is finite, I must show that all singletons are open. By contrapositive, $\{\{x\}|x\in X\}$ forms an open cover without finite subcover hence $X$ is not compact iff it is not finite.
How to resolve this chicken-egg problem?!
Thanks!
Recall that Hausdorff spaces are $T_1$, but also that compact subsets of Hausdorff spaces are closed.