The question is stated as:
Prove: $(\forall B)(B \in F \Rightarrow C \subseteq B) \Rightarrow C \subseteq \bigcap_{A \in F}A$
Thats what i thinked in a textual way: If we assume that for every $B$, if it belongs to $F$ then $C$ is a subset of $B$, we have that all elements from $C$ are in every set of $F$, thus the intersection of $F$ contains at least the elements of $C$, and then $C \subseteq \bigcap F$.
I tried to do something with symbols but i got stucked in the second line:
$$(\forall B)(B \in F \Rightarrow C \subseteq B)$$ $$(\forall B)(B \in F \Rightarrow (\forall x)(x \in C \Rightarrow x \in B))$$ $$(\forall B)(B \notin F \lor (\forall x)(x \notin C \lor x \in B))$$
I dont know how to proceed, so i want to know if the textual part is correct, how this proof should be done and if its possible to do some operation with the quantifiers.
Assume that the LHS holds. We want to show the RHS, i.e. if $x\in C$, then $x\in A$ for all $A\in F$.
For this, let us take $x\in C$ an fix some $A\in F$. The LHS now implies that $x\in C\subset A$, so $x\in A$. As this holds for arbitrary $A\in F$, we are done.