real analysis - Proving the pre-image of this set

Question:
Let $f: \left ( X,d \right )\rightarrow \left ( Y,e \right )$ be a map between metric spaces.
Let $U\subset X$ and let $V \subseteq Y.$

Show that if $V'\subseteq V$ then $f^{-1}\left ( V' \right )\subseteq f^{-1}\left ( V \right ).$

By the definition of pre-image:

$f^{-1}\left ( V' \right )=\left \{ y \in Y \mid f\left ( y \right ) \in V'\right \}.$

Useful hints are appreciated.
Thanks in advance.

Answer

We will show that any element $x\in f^{-1}\left ( V' \right )$ is also an element of $f^{-1}\left ( V \right )$.

Let $x\in f^{-1}\left ( V' \right )$ then, by definition of preimage, there is $y\in V'$ such that $f(x)=y$.

Now $V'\subseteq V$ implies that $y\in V$ and therefore $x\in f^{-1}\left ( V \right )$.