Prove that if $\forall A \in \mathcal F (B\subseteq A)$ then $B \subseteq \bigcap \mathcal F $ This is Velleman's exercise 3.3.10. Suppose that $\mathcal F$ is a nonempty family of sets, B is a set, and $\forall A \in \mathcal F (B\subseteq A)$. Prove that $B \subseteq \bigcap \mathcal F $.
My approach so far:
Suppose that $A \in \mathcal F$, suppose that $x \in B$. Then $x$ is an element of any $A \in \mathcal F$ because  $\forall A \in \mathcal F (B\subseteq A)$. Then $x$ will also be an element of $\bigcap \mathcal F$ because $\forall A \in\mathcal F(x \in A)$. Hence, any element of $B$ is an element of $\bigcap \mathcal F$. This shows that $B \subseteq \bigcap \mathcal F$.
Is this a valid proof? Thanks in advance.

Basically yes, but you should try to be a bit more methodical to obtain a 100% correct proof. Here are the steps to take (compare with what you've actually done). 
1) Suppose $x\in B$. The aim is to show that $x\in \bigcap \mathcal F$. 
2) To show $x$ is in an intersection, it suffices to show that $x\in A$ for all $A\in \mathcal F$. 
3) So, let $A\in \mathcal F$. 
4) Now finish the proof. 
You will notice that the first three steps are mechanical. They do not require any thinking. This is the part you got a bit wrong in your proof. 
