Show every irreducible subset of a topological space $X$ is contained in a maximal irreducible subset

Suppose $\{A_i\}_{i \in I}$ is a chain of irreducible subsets of $X$. Let's try to show that $C = \bigcup_{i \in I} A_i$ is irreducible. So let $U,V$ be non-empty open subsets in $C$. Then $U \cap A_i$ and $V \cap A_j$ are also non-empty for some $i,j \in I$. As we have a chain, either $A_i \subseteq A_j$ or $A_j \subseteq A_i$, say the former. Then $U \cap A_j,V \cap A_j$ are non-empty open subsets of $A_j$, so they intersect. Thus $U, V$ also intersect, so $\bigcup_i A_i$ is irreducible.

Now we can apply Zorn's lemma to the poset of all irreducible subsets of $X$ that contains some fixed $A$, ordered by inclusion, as the union is clearly an upperbound for the chain. This gives us a maximal element around every irreducible subset $A$. (As $A$ itself is in this poset, it's non-empty).

First, let me note that there might be more than one maximal irreducible set containing $A$. For instance, let $X=\{a,b,c\}$ with $\{b\}$ and $\{c\}$ as the only nontrivial open sets. Then $A=\{a\}$ is irreducible, but $\{a,b\}$ and $\{a,c\}$ are two different maximal irreducible sets containing it.

In particular, in this example, your $\hat{A}$ would be all of $X$, which is not irreducible. So your approach will not work. More generally, you should not expect there to be any canonical way to construct a maximal irreducible set, because of this non-uniqueness.

So instead, you need to do something nonconstructive. I would suggest trying out Zorn's lemma.