Union of Open Irreducible Non-Disjoint Subspaces is Irreducible

Theorem
Let $T = \struct {S, \tau}$ be an irreducible toplogical space.

Let $U$ and $V$ be open irreducible subspaces.

Let their intersection $U \cap V$ be non-empty.

Then their union $U \cup V$ is irreducible.

Proof
In view of the definition of the subspace topology, it suffices to show that for all closed sets $A_1$ and $A_2$:
 * $U \cup V \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : U \cup V \subseteq A_{i_0}$

To this end, let $A_1$ and $A_2$ closed sets such that:
 * $U \cup V \subseteq A_1 \cup A_2$.

Then, in particular:
 * $U \subseteq A_1 \cup A_2$.

Since $U$ is irreducible, we have:
 * $(1):\quad \exists i_0 \in \set {1,2} : U \subseteq A_{i_0}$

Then:

Thus:
 * $V = \paren {V \setminus U} \cup \paren {V \cap A_{i_0} }$

Here, both $V \setminus U$ and $V \cap A_{i_0}$ are closed in $V$ with respect to subspace topology.

As $U \cap V$ is non-empty, we have:
 * $V \setminus U \subsetneq V$

That is, $V \setminus U$ is a proper subset of $V$.

Since $V$ is irreducible, $V \cap A_{i_0}$ is not a proper subset of $V$.

That is:
 * $V = V \cap A_{i_0}$

Hence:
 * $V \subseteq A_{i_0}$

Together with $(1)$, we conclude:
 * $U \cup V \subseteq A_{i_0}$