TITLE: Show every irreducible subset of a topological space $X$ is contained in a maximal irreducible subset
QUESTION [5 upvotes]: Let $X$ be a topological space. A subset $A$ is irreducible if for every open $U,V\subseteq A$, we have $U\cap V\neq\varnothing$. Show that any irreducible subset $A\subseteq X$ is contained in a maximal irreducible set.

So here's basically what I want to do: let $A$ be an irreducible subset of $X$ and $\hat A$ be the union of all irreducible subsets containing $A$. I think this is the maximal irreducible subset I'm looking for. 
To show this, let $U,V\subseteq$ be open in $A$. I want to show $U\cap V\neq\varnothing$ but I'm not sure how to do this. It's clear that an open subset of an irreducible set is irreducible, so I could show this if I knew that $U$ and $V$ were both contained in an irreducible set. But I don't know if this is even true. Any hints?

REPLY [4 votes]: 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).