Space admitting an irreducible connected open covering is irreducible

Question

Let $\{U_i:i \in I\}$ be an open covering of topological space $X$, where $U_i \cap U_j \neq \varnothing$ for every $i,j$. If $U_i$ is irreducible for all $i \in I$ then $X$ is irreducible.

I am stuck as to how to show this, any help is appreciated!

Answer 1

Suppose $X$ is not irreducible, consider $V_i$ the adherence of $U_i$ it is irreducible. So there exists $x\in X$ not in $V_i$, $x\in U_j$, let $W_i$ be the complementary space of $V_i$, $U_j=U_j\cap W_i\bigcup U_j\cap U_i$, since $U_i\cap U_j$ is not empty and $U_j$ is irreducible, we deduce that $W_i\cap U_j$ is empty. Contradiction, since $x\in W_i\cap U_j$.