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

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.
