Proposition 3

(i) If $X$ is an irreducible space, every non-empty open subset of $X$ is irreducible.

(ii) Let $(U_\alpha)_{\alpha \in A}$ be a non-empty covering of a topological space $X$ consisting of open sets such that $U_\alpha \cap U_\beta = \emptyset$ for every pair of indices $(\alpha, \beta)$. If the sets $U_\alpha$ are irreducible, the space $X$ is irreducible.

(i) If $X$ is irreducible, $U \subset X$ non-empty and open in $X$ and $V \subset U$ non-empty and open in $U$, $V$ is also open in $X$, therefore dense in $X$ and a fortiori dense in $U$. Then $U$ is irreducible (Proposition 1).

(ii) Let us show that, for every non-empty open set $V$ in $X$, $V \cap U_\alpha \neq \emptyset$ for all $\alpha \in A$: it follows that $V \cap U_\alpha$ is dense in $U_\alpha$, by hypothesis, hence that $V$ is dense in $X$ and this proves that $X$ is irreducible (Proposition 1). Now there exists at least one index $y$ such that $V \cap U_y \neq \emptyset$; as $U_\alpha \cap U_y \neq \emptyset$ for all $\alpha$ and $V \cap U_y$ is dense in $U_y$, $U_y \cap V \neq \emptyset$ and a fortiori $U_y \cap V \neq \emptyset$, which completes the proof of (ii).