The question
Suppose $F$ and $G$ are non-empty families of sets. Prove if $F \subseteq G$, then $ \cap G \subseteq \cap F$.
My proof
Let $x$ be arbitrary element of $F$. Let $y$ be arbitrary element of $\cap G$. Assume $y \in \cap G$. This means for all arbitrary set $A$ in $G$, $y$ is element of $A$. Let $B$ be arbitrary set in $F$ and assume $B \in F$. Since $B$ is an element of $F$ and $F$ is a subset of $G$, $B$ is an element of $G$. If $y$ is element of $\cap G$, then $y$ is an element of $B$ since $B$ is element of $G$. Therefore, if $y \in \cap G$ then $y\in \cap F$, which is what $\cap G \subseteq \cap F$ means.
Is the proof accurate? What can I do to make the proof more clear?
Thank you for your time