Let X,Y be sets, and let $f:X\rightarrow Y$ be a function. Prove:
Let $x \in f(f^{-1}(B))$. Then there is a $u \in f^{-1}(B)$ such that $f(u) = x$. Since $u \in f^{-1}(B)$, by definition, we must have that $f(u) \in B$. But recall that $x = f(u)$, so $x \in B$.
Recall the definitions: $f^{-1}(B)=\{x\in X\mid f(x)\in B\}$ and $f(A)=\{f(x)\mid x\in A\}$. Thus $$\begin{align}& y\in f(f^{-1}(B))\\\iff &y=f(x)\text{ for some }x\in f^{-1}(B)\\\iff &y=f(x)\text{ for some }x\text{ with }f(x)\in B\\ \implies&y\in B\end{align}$$
Assume $f$ is surjective and $B\subset Y$. Then there exists (by surjectivity) some $x\in X$ with $f(x)=y$ for every $y\in Y$, especially for every $y\in B$. This $x$ is in $ f^{-1}(B)$, thus showing that $y=f(x)\in f(f^{-1}(B))$.
On the other hand, assum ethat $f$ is not surjective, say $y\notin f(X)$. Then $f(f^{-1}(\{y\}))=f(\emptyset)=\emptyset\ne\{y\}$.
With $f : X \rightarrow Y$, $x \in X$, $y \in Y$, $A \subset X$, and $B \subset Y$ assumed throughout, and using the definitions $$ \begin{align} (0) & y \in f(A) \;\equiv\; \langle \exists x : x \in A : f(x) = y \rangle \\ (1) & x \in f^{-1}(B) \;\equiv\; f(x) \in B \\ \end{align} $$ the simplest thing I see is to calculate which elements are in $f(f^{-1}(B))$, since that is the most complex expression in the question: for every $y$ and $B$, we have $$ \begin{align} & y \in f(f^{-1}(B)) \\ \equiv & \;\;\;\;\;\text{"expand using definition (0)"} \\ & \langle \exists x : x \in f^{-1}(B) : f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"expand using definition (1)"} \\ & \langle \exists x : f(x) \in B : f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"logic: substitute $y$ for $f(x)$"} \\ & \langle \exists x : y \in B : f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"logic: simplify, allowed because expression $y \in B$ does not contain $x$"} \\ & y \in B \;\land\; \langle \exists x :: f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"make $x \in X$ explicit -- to prepare for use of definition (0)"} \\ & y \in B \;\land\; \langle \exists x : x \in X: f(x) = y \rangle \\ \equiv & \;\;\;\;\;\text{"introduce definition (0)"} \\ & y \in B \;\land\; y \in f(X) \\ \equiv & \;\;\;\;\;\text{"definition of $\cap$"} \\ & y \in B \cap f(X) \\ \end{align} $$ (For clarity I've left in a few trivial steps that one would normally skip.) So by set extensionality we've calculated that $$ (*) \;\;\;\;\; f(f^{-1}(B)) \;=\; B \cap f(X) $$ From this key insight, both parts of the question follow directly. For the first part use $B \cap f(X) \subset B$ from set theory. For the second part, surjectivity automatically falls out of the following calculation: $$ \begin{align} & \langle \forall B :: f(f^{-1}(B)) = B \rangle \\ \equiv & \;\;\;\;\;\text{"using (*)"} \\ & \langle \forall B :: B \cap f(X) = B \rangle \\ \equiv & \;\;\;\;\;\text{"make range explicit; set theory"} \\ & \langle \forall B : B \subset Y : B \subset f(X) \rangle \\ \equiv & \;\;\;\;\;\text{"set theory"} \\ & Y \subset f(X) \\ \equiv & \;\;\;\;\;\text{"set extensionality -- so that we can use our definitions again"} \\ & \langle \forall y : y \in Y : y \in f(X) \rangle \\ \equiv & \;\;\;\;\;\text{"introduce definition (0)"} \\ & \langle \forall y : y \in Y : \langle \exists x : x \in X: f(x) = y \rangle \rangle \\ \equiv & \;\;\;\;\;\text{"definition of surjectivity"} \\ & f \textrm { is surjective} \\ \end{align} $$
