The Stacks project

Proof. Suppose $X = Z_1 \cup Z_2$ with $Z_ i$ closed. Consider the open sets $U_1 = Z_1 \setminus Z_2 = X \setminus Z_2$ and $U_2 = Z_2 \setminus Z_1 = X \setminus Z_1$. To get a contradiction assume that $U_1$ and $U_2$ are both nonempty. By (b) we see that $f(U_ i)$ is open. By (a) we have $Y$ irreducible and hence $f(U_1) \cap f(U_2) \not= \emptyset$. By (c) there is a point $y$ which corresponds to a point of this intersection such that the fibre $X_ y = f^{-1}(y)$ is irreducible. Then $X_ y \cap U_1$ and $X_ y \cap U_2$ are nonempty disjoint open subsets of $X_ y$ which is a contradiction. $\square$