A weak kind of fixed point Let $X$ be a set and let $\cal A$ be a non-empty subset of $P(X)$ with the property that whenever $A_1 \subseteq A_2 \subseteq \cdots $ is an increasing chain of elements of $\cal A$ then $\cup_i A_i \in \cal A$. Let 
$f : \cal A\longrightarrow \mathbb{R}$ be increasing and bounded from above. Is it true that there must exists $M\in\cal A$ such that for any element $C$ of $\cal A$ containing $M$, $f(M)= f(C)$ ?  
 A: If not, then we can find an increasing and continuous sequence $(A_\alpha: \alpha < \omega_1)$ of elements of $\mathcal{A}$ such that $\alpha < \beta \implies f(A_\alpha) < f(A_\beta)$. This is a contradiction.

Edit. Let me add more details. By contradiction, the following holds:
$(*)$: For any $M \in \mathcal{A}$, there exists $M \subset C \in \mathcal{A}$ with $f(M) < f(C).$
Let $A_0\in \mathcal{A}$ be arbitrary. Given $A_\alpha \in \mathcal{A},$ use $(*)$ to find $A_{\alpha+1} \in \mathcal{A}$ with $A_\alpha \subset A_{\alpha+1}$ and $f(A_\alpha) < f(A_{\alpha+1}).$
Now suppose $\alpha<\omega_1$ is limit and we have defined $(A_\beta: \beta< \alpha)$ which is increasing and continuous and for all $\beta_1 < \beta_2 < \alpha, f(A_{\beta_1}) < f(A_{\beta_2}).$ Let $A_\alpha=\bigcup_{\beta<\alpha}A_\beta.$ Note that $A=\bigcup_{n<\omega}A_{\beta_n},$ where $(\beta_n: n<\omega)$ is increasing cofinal in $\alpha,$ so by assumption $A_\alpha \in \mathcal{A}.$
It is also clear that for all $\beta<\alpha, f(A_\beta) < f(A_\alpha)$ (as $f$ is increasing by assumption).
But this is impossible, as otherwise we can find $q_\alpha \in \mathbb{Q} \cap (f(A_\alpha), f(A_{\alpha+1})), \alpha < \omega_1,$ and for $\alpha \neq \beta, q_\alpha \neq q_\beta,$ so $\mathbb{Q}$ is uncountable which is impossible. 
A: Yes. Denote $g(A)=\sup_{B\in {\mathcal A},A\subset B} f(B)$. This function decreases. Start with some $A_1\in {\mathcal A}$. Choose $A_2$ so that $A_1\subset A_2$ and $f(A_2)>g(A_1)-1$. On $n$-th step choose $A_{n+1}\supset A_n$ so that $f(A_{n+1})>g(A_n)-2^{-n}$. We have $f(A_n)\leqslant f(A_{n+1})\leqslant g(A_{n+1})\leqslant g(A_n)$ and $g(A_{n+1})-f(A_{n+1})\leqslant g(A_n)-f(A_{n+1})<2^{-n}$. Thus there exists a common limit $c=\lim f(A_n)=\lim g(A_n)$ and $c\geqslant g(\cup A_i)\geqslant f(\cup A_i)\geqslant c$ since $f$ increases and $g$ decreases, so $\cup A_i$ serves as your $M$. 
