If $f^{-1}(E)$ is empty, then result R7: Empty set is subset of every set establishes the claim so that we may assume that $f^{-1}(E)$ is nonempty. If \begin{equation} x \in f^{-1}(E) = \{ z \in X : f(z) \in E \} \end{equation} then $f(x) \in E$. Since $E$ is contained in $F$, then the transitivity of the subset relation guarantees that $f(x) \in F$ and therefore $x \in f^{-1}(E)$. Since $x \in f^{-1}(E)$ was arbitrary, we have the inclusion $f^{-1}(E) \subseteq f^{-1}(F)$,. $\square$