The book I'm using enunciates:
$\textbf{Theorem:}$ Let X be set and $\sigma:P(X) \to P(X)$ increasing function $(x_1 \subset x_2 \subset X \Rightarrow \sigma(x_1) \subset \sigma(x_2) \subset X)$ then $\sigma$ has fixed point.
$\textbf{Proof:}$ Consider $\tau= \{ x_1 \in P(X): x_1 \subset \sigma(x_1) \} \subset P(X)$ and take $x_0 := \bigcup \tau = \bigcup \{ x_1 \in P(X): x_1 \subset \sigma(x_1) \} $ then $\sigma(x_0)=x_0$.
I can not showed $\sigma(x_0)=x_0$.