Suppose that $f:\mathcal{P}(A) \to \mathcal{P}(A)$ is an increasing mapping with respect to $\subseteq$, i.e. if $B\subseteq C$ then $f(B)\subseteq f(C)$. Then there exists $G\subseteq A$ such that $f(G)=G$.
My attempt:
Let $\mathcal G=\{B\in \mathcal{P}(A) \mid B\subseteq f(B)\}$ and $G=\bigcup\limits_{B\in\mathcal G}B$. If $B\in\mathcal G$ then $B\subseteq G$ and consequently $f(B)\subseteq f(G)$. Thus $B\subseteq f(B)\subseteq f(G)$ for all $B\in\mathcal G$ and thus $G=\bigcup\limits_{B\in\mathcal G}B \subseteq f(G)$. Hence $G\in\mathcal G$.
Since $G\subseteq f(G),f(G)\subseteq f(f(G))$. As a result, $f(G)\in \mathcal G$. It follows that $f(G)\subseteq G$.
To sum up, we have $G\subseteq f(G)$ and $f(G)\subseteq G$. This implies that $f(G)=G$.
Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!