TITLE: Connected and irreducible topological spaces
QUESTION [15 upvotes]: A topological space is called connected if any presentation of $X$ as $X = V_1 \uplus V_2$ by disjoint open subsets implies that one of them is trivial ($V_1 = X$ or $V_2 = X$). By taking complement one can replace the word "open" by "close".
A topological space is called irreducible if any presentation of $X$ as $X = X_1 \cup X_2$ by two closed subsets implies that one of them is trivial ($X_1 = X$ or $X_2 = X$).
Clearly every irreducible space is connected. The converse is not always true but:

Proposition:
  Let $X$ be a connected topological space which has an open covering by irreducible subspaces. Then $X$ is irreducible.

I want to prove this proposition.

REPLY [9 votes]: HINT: It’s easily checked that a space $X$ is irreducible if and only if $U\cap V\ne\varnothing$ whenever $U$ and $V$ are non-empty open subsets of $X$. Let $X$ be connected, and let $\mathscr{U}$ be an open cover of $X$ by irreducible subspaces. Suppose that $V$ and $W$ are disjoint open subsets of $X$; you want to show that one of them is empty.

Show that each $U\in\mathscr{U}$ intersects at most one of $V$ and $W$.

Let 
$$\begin{align*}
&\mathscr{U}_V=\{U\in\mathscr{U}:U\cap V\ne\varnothing\}\;,\\
&\mathscr{U}_W=\{U\in\mathscr{U}:U\cap W\ne\varnothing\}\;,\text{ and}\\
&\mathscr{U}_0=\mathscr{U}\setminus(\mathscr{U}_V\cup\mathscr{U}_W)\;.
\end{align*}$$
Consider the open sets $\bigcup\mathscr{U}_V,\bigcup\mathscr{U}_W$, and $\bigcup\mathscr{U}_0$, bearing in mind that $X$ is connected.