If $A \subseteq X\land B \subseteq Y$ are any sets, prove that $f(A\cap f^{-1}(B)) \subseteq f(A) \cap B$ 
If $A \subseteq X\land B \subseteq Y$ are any sets, and $f:X\to Y$, prove that $f(A\cap f^{-1}(B)) \subseteq f(A) \cap B$

Here is what I've done for the proof, I just need a little bit of guidance in finishing it up.
Proof: 
Suppose $A \subseteq X \text{ and } B \subseteq Y$, Let $z \in f(A \cap f^{-1}(B))$ be arb.
First I unpacked the goal statement:
$ \Rightarrow z \in f(A) \cap B $
$ \Rightarrow z \in f(A) \land z \in B $
$ \Rightarrow \exists x \in A \text{ s.t } z = f(x) \land z \in B $
Next I unpacked the assumptions:
$
\Rightarrow \exists x \in A \cap f^{-1}(B) \text{ s.t } z = f(x)
$
$
\Rightarrow \exists x \in A \land x \in f^{-1}(B) \text{ s.t } z = f(x)
$
$
\Rightarrow \exists (x \in A \text{ s.t } z = f(x)) \land (x \in Y \land x \in B \text{ s.t } z = f(x))
$
So I've proven that $ \exists x \in A \text{ s.t } z = f(x) $, but how do I go about proving that $x \subseteq Y$ and $z \in B$?

In fact, wa have equality. For the case :
$$f(A) \cap B \subset f \left(A \cap f^{-1}(B)\right)$$
Let $y \in f(A) \cap B$ then $y \in f(A)$ and $y \in B$.
We have $y \in f(A)$ then :
$$\exists x \in A, y = f(x)$$
We deduce that $f(x) = y \in B$ then $x \in f^{-1}(B)$.
We have $x \in A$ and $x \in f^{-1}(B)$ then :
$$x \in A \cap f^{-1}(B)$$
Hence :
$$y = f(x) \in f \left(A \cap f^{-1}(B)\right)$$
Finally :
$$f(A) \cap B \subset f \left(A \cap f^{-1}(B)\right)$$
