Why does a zero-dimensional irreducible space have no non-trivial open subsets?

Question

Let $X$ be an irreducible topological space. Define the dimension to be the supremum of the length of chains, $$ Z_{0} \supset Z_{1} \supset \cdots \supset Z_{n} $$ of distinct, irreducible closed subsets. If $X$ has dimension $0$, why are the only open sets itself and the empty set? I feel like I am very close, but I just can't find a nice proof for it. Say there was a non-trivial open subset $U \subseteq X$. Let $Y$ be its complement. I would like to claim that $Y$ would be an irreducible closed subset which would create a chain of lenth $1$. Is somebody able to spell out how to do this?

Answer 1

If $U\subset X$ is open, then the complement $Z = X\setminus U$ may not be irreducible.

However, if $p\in Z$ is any point, then the closure $\overline{\{p\}} \subset Z$ is an irreducible closed set. Therefore, if $X$ is irreducible of dimension zero, we must have $Z=\emptyset$ or $Z=X$.