# Prove that $f(A) \subseteq B \implies A \subseteq f^{-1}(B)$

Let $A,B ⊆ \mathbb R \$ and $f\colon\mathbb R \to \mathbb R$ be a function. Then $f(A) \subseteq B \implies A \subseteq f^{-1}(B)$.

Take any $a\in A$. Since $f(a)\in f(A)\subseteq B$, we have $a\in f^{-1}(B)$.
You do not need the continuity - only the fact that $f$ is a function $f:X\to X$ with $A,B\subseteq X$.
Take any $x \in A$. We know that $y = f(x) \in B$. Therefore there exists $y \in B$ such that $y = f(x).$ This is the definition of $f^{-1}(B)$. So $x \in f^{-1}(B)$.
Note that this holds for any set $A$ and $B$ and any function $f$ as long as $f(A) \subseteq B$. (No need for continuity.)