Proposition. Let $X$ be a topological space and $U\subseteq X$ an open set of $X$. If $Z$ is a closed irreducible set of $X$ that meets $U$ then $Z\cap U$ is a closed irreducible set of $U$.
Note. Irreducibility is an intrinsic property that does not depend on ambient space (irreducible in $U$ is the same as irreducible in $X$)
I'm having trouble writing a proof of this: $Z\cap U$ is clearly a closed set of $U$ but none of the equivalent definitions of irreucibility that i know of seem to work, for example if $Z\cap U=F\cup G$ with $F,G$ closed in $Z\cap U$ then $Z\cap U=(Z\cap U)\cap (F^*\cup G^*)$ for some $F^*,G^*$ closed in $Z$ and thus in $X$. I don't know how to continue from here, no rearranging of the last expression works. Why does $U$ need to be open?