I'm trying to show that the following statement is false:
Let $x$ be a non-empty set. Let $F$ be a proper countably complete filter on $X$. Then there is a countably complete proper ultra filter $H$ on $X$ s.t. $F \subset H$.
Countably complete means the following: $ \forall H \subset F (H \text{ is countable } \rightarrow \bigcap H \in F)$ $F$ is called proper if $\varnothing \notin F$.
My idea was to pick an uncountable set $x$ and let $F = \{ y \subset x \mid y^c \text{ is countable } \}$ be the cocountable filter. Then pick any $y \in F$ so that $y^c$ is countable. Let's denote it $y^c = \{ y_n \mid n \in \mathbb N_{>0} \}$. Let $(P, \le)$ be the set of all proper countably complete filters $G$ containing $F$. I wanted to make a chain extending $F$. But I can't seem to make it work. Something like $C_0 = F$ and either $C_n = flt(C_{n-1}, \{y_1, \dots, y_n \})$ or $C_n = flt(C_{n-1}, y^c \setminus \{y_1, \dots, y_n \})$ where $flt(F,s) = \{ w \subset x \mid \exists z \in F ( z \cap s \subset w \}$ doesn't work. How can I construct a chain such that $\bigcup \mathcal C$ is not closed with respect to countable intersection?
Thanks for your help.
Edit
Background information: I've done parts (a) and (b) of the following exercise and this is part (c):