When do we have $A \subset B$ imples $f^{-1}(A) \subset f^{-1}(B)$? Im not sure if continuity of the function $f$ is enough to have the above. Or is it the monotonicity of $f$?

Here is an alternative proof, where we start with the most complex side and expand the definitions, and then work towards $\;A \subseteq B\;$:
\begin{align}
& f^{-1}[A] \;\subseteq\; f^{-1}[B] \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \langle \forall x :: x \in f^{-1}[A] \;\Rightarrow\; x \in f^{-1}[B] \rangle \\
\equiv & \qquad \text{"basic property of $\;^{-1}[\quad]\;$, twice"} \\
& \langle \forall x :: f(x) \in A \;\Rightarrow f(x) \in B \rangle \\
\Leftarrow & \qquad \text{"logic: generalize $\;f(x)\;$ to any $\;y\;$ -- to get rid of $\;f\;$"} \\
& \langle \forall y :: y \in A \;\Rightarrow y \in B \rangle \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& A \subseteq B
\end{align}
