$f:X \rightarrow Y$ where $A \subset X$ and $B \subset Y$
Prove (if it is correct) that $$B \subset f(f^{-1}(B))$$ This is my approach
Proof
Choose an arbitrary element $y \in B$, by the definition of inverse image, we know that $x \in f^{-1}(B)$ such that $f(x)=y$. As $y \in B$ and $x \in f^{-1}(B)$, we can conclude that $y \in f(f^{-1}(B))$.
Is this correct?
This is false. Let $A = X = \{0\}$, $Y = \{0, 1\}$ and $B = \{1\}$. Let $f : X \to Y$ be the identity function. Then $f^{-1}(B) = \emptyset$, so $f(f^{-1}(B)) = \emptyset$ but $B$ is not empty.
The error in your reasoning is the statement "...by the definition of inverse image, we know that $x \in f^{-1}(B)$ such that $f(x) = y$". This is not true because the definition of inverse image states that: $$ f^{-1}(B) = \{x \in X \mid f(x) \in B\} $$ This definition asserts that $f$ maps every element in $f^{-1}(B)$ to $B$, but not necessarily to every element in $B$.
As such, if you assume that $f$ is surjective, then your statement holds - you can safely conclude that such an $x$ which $f(x) = y$ exists.