You really have three things to prove:
- $f(A\cap B)\subseteq f(A)\cap f(B)$;
- $f^{-1}(C\cap D)\subseteq f^{-1}(C)\cap f^{-1}(D)$; and
- $f^{-1}(C)\cap f^{-1}(D)\subseteq f^{-1}(C\cap D)$.
(The last two together of course give you $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$.) Each of these can be done by ‘element-chasing’. To get you started, I’ll illustrate with the first.
Suppose that $y\in f(A\cap B)$. By definition this means that there is some $x\in A\cap B)$ such that $y=f(x)$. By the definition of intersection, $x\in A$ and $x\in B$. Since $x\in A$, by definition $y=f(x)\in f(A)$. Similarly, $y\in f(B)$, and it follows from the definition of intersection that $y\in f(A)\cap f(B)$. Since $y$ was an arbitrary element of $f(A\cap B)$, this shows that every element of $f(A\cap B)$ belongs to $f(A)\cap f(B)$, i.e., that $f(A\cap B)\subseteq f(A)\cap f(B)$.
You can use the same kind of argument to prove each of the other two inclusions.
I’ve included more detail in the justifications than is perhaps strictly necessary, even at a beginning stage, but in general it’s best at first to include more explanation rather than less. There are two reasons for this. First, it forces you to think about the precise justifications for your statements; this helps to avoid one of the most common sources of careless error $-$ slipping in something that seems intuitively obvious without checking that it’s actually provable. Secondly, it makes it much easier for your instructor to tell whether you actually understand what you’re doing and to clear up any misconceptions that you may have.
Added: The definitions of image and preimage that you’re using are fine, though I don’t really like the notation that they use. I would write $$f[A]=\{f(a):a\in A\}$$ and $$f^{-1}[A]=\{x\in X:f(x)\in A\}\;,$$ but the substance of the definitions is the same. The image of a set $A\subseteq X$ under $f$ is the set of all ‘outputs’ of $f$ when the ‘inputs’ come only from $A$; the preimage of a set $A\subseteq Y$ under $f$ is the set of all points in $X$ that $f$ maps to points in $A$.
Since I dealt only with images in my original answer, perhaps I should say a little more about preimages. Suppose that $x\in f^{-1}(A)$ for some set $A\subseteq Y$. This means that $x$ is one of the points of $X$ taken by $f$ to some point of $A$. In other words, $f(x)$ must be in $A$. Conversely, if $f(x)\in A$, then by definition $x\in f^{-1}(A)$. Thus, $x\in f^{-1}(A)$ if and only if $f(x)\in A$. With that in hand, let’s look at the third of the three things that I listed above.
Suppose that $x\in f^{-1}(C)\cap f^{-1}(D)$; by definition of intersection this means that $x\in f^{-1}(C)$ and $x\in f^{-1}(D)$. We’ve just seen that $x\in f^{-1}(C)$ if and only if $f(x)\in C$. Similarly, $x\in f^{-1}(D)$ if and only if $f(x)\in D$. Thus, we now know that $f(x)\in C$ and $f(x)\in D$, i.e., that $f(x)\in C\cap D$. But this means that $x\in f^{-1}(C\cap D)$, and since $x$ was an arbitrary element of $f^{-1}(C)\cap f^{-1}(D)$, it follows that $f^{-1}(C)\cap f^{-1}(D)\subseteq f^{-1}(C\cap D)$.
"Again, take any subset $X \subseteq A \cap B$."
You might want to change the wording here. "Take any set $X$ such that $X \subseteq A$ and $X \subseteq B$. Then $X \subseteq A \cap B$."
As written it sounds like you're assuming what you want to prove. Otherwise good...
Best Answer
The comment at your edit is true.
Suppose that $f(x) = x^{2}$, $A = [-1,0]$ and $B = [0,1]$.
Then we have that $f(A\cap B) = \{0\}$ and $f(A)\cap f(B) = [0,1]$.
Consequently, $f(A)\cap f(B)\not\subseteq f(A\cap B)$.
On the other hand, if $f$ is injective, then we deduce that \begin{align*} y\in f(A)\cap f(B) & \Rightarrow (y\in f(A))\wedge(y\in f(B))\\\\ & \Rightarrow (\exists a\in A)(y = f(a))\wedge(\exists b\in B)(y = f(b))\\\\ & \Rightarrow y = f(a) = f(b)\\\\ & \Rightarrow a = b\in A\cap B\\\\ & \Rightarrow y\in f(A\cap B) \end{align*}
and we are done.
Hopefully this helps.