Open Set of Irreducible Space is Irreducible

Theorem
Let $X$ be an irreducible topological space.

Let $U$ be a non-empty open subset of $X$.

Then $U$ is irreducible in its induced subspace topology.

Proof
Suppose $U$ is reducible. Then $U=V_1\cup V_2$ for some closed subsets $V_1$ and $V_2$ of $U$. Now, by definition of subspace topology, it follows that $V_1=U\cap W_1$ and $V_2=U\cap W_2$ for some closed subsets $W_1$ and $W_2$ of $X$. Now, it is clear that $W_1\neq X$, because $W_1=X$ $\implies$ $V_1=U\cap W_1=U$. Contradiction, because $V_1$ is a proper subset of $U$. Now, $U\neq\emptyset$ implies that $X-U$ is a proper closed subset of $X$. Let $W_3:=W_2\cup(X-U)$. Then $W_3$ is a closed subset of $X$. Also, $W_3\neq X$, because otherwise $V_2=U\cap W_2=U\cap W_3=U$. And so, $X=W_1\cup W_3$ shows that $X$ is reducible. Contradiction! Q.E.D.

Also see

 * Space is Irreducible iff Open Subsets are Connected
 * Closed Subset of Ultraconnected Space is Ultraconnected, whose proof is almost the same