$f[A]\cap f[B]\supsetneq f[A\cap B]$ - Where does the string of equivalences fail ? [Chartrand 3E 9.12(b), 9.29] I only realised that equality may fail in $f[A]\cap f[B]\supseteq f[A\cap B]$
 (i.e., that we can have $A,B,f$ for which $f[A]\cap f[B]\neq f[A\cap B]$) after checking the answer. 
I don't see any problems in this spurious string of equivalences which seems to imply that $f[A]\cap f[B]$ is always equal to $f[A\cap B]$. So where does it fail?   
I'm not enquiring  about counterexamples for the failure of $f[A]\cap f[B]\subseteq f[A\cap B]$. 
I'm also aware of 9.29: If $f$ is a one-to-one function, then $f[A]\cap f[B] = f[A\cap B]$. Where can this be applied to fix the broken string? Please advise me of other problems. 

$\begin{align}
\text{ A function } f(x)\in f(A\cap B)
&\iff x\in A\cap B\\
&\iff x\in A \; \wedge \; x\in B\\
&\iff f(x)\in f(A) \; \wedge \; f(x)\in f(B)\\
&\iff f(x)\in f(A) \cap f(B)
\end{align}$

 A: It is always the case that $f[A\cap B]\subseteq f[A]\cap f[B]$. If $y\in f[A\cap B]$ then there is an $x\in A\cap B$ such that $f(x)=y$.  It follows that $f(x)\in f[A]$ and $f(x)\in f[B]$. 
However, it is not always the case that $f[A\cap B]\supseteq f[A]\cap f[B]$. If we look at your purported proof, the error is in the first inference (working from the bottom to the top): Suppose that $y \in f[A]\cap f[B]$.  Then there exist $x_1\in A$ and $x_2\in B$ such that $f(x_1)=f(x_2)=y$. However, unless $f$ is injective, you cannot assume that $x_1=x_2$. 

There's another error too:
$f(x)\in f[A\cap B]$ is not equivalent to requiring that $x\in A\cap B$.  Again, this condition only holds for all $x$ in the case that $f$ is an injection. 
A: Let 
$$f:x\mapsto x^2$$
and
$$A=(0,\infty)\quad;\quad B=(-\infty,0)$$
then we have
$$f(A\cap B)=\emptyset\subset f(A)\cap f(B)=(0,\infty)$$
Edit In your proof you have two mistakes:


*

*The first by writing $$f(x)\in f(A\cap B)\Rightarrow x\in A\cap B$$
but  in my counterexample we have $f(-1)=1\in f(A)$ and $-1\not\in A=(0,\infty)$.

*The second by writing
$$f(x)\in f(A)\land f(x)\in f(B)\Rightarrow x\in A\land x\in B$$
and the counterexample is
$1=f(1)\in f(A)\land f(1)\in f(B)$ but $1\not\in B=(-\infty,0)$

A: Your problem is that $s \in S$ is not equivalent to $g(s) \in g[S]$. You have made this mistake several times in your derivation.
As a simple counter-example, let $g$ be the function $g(x) = x^2$ on the reals, let $S$ be the set of positive numbers, and let $s = -1$.
A: The proof above actually shows that $f(A \cap B) \subseteq f(A) \cap f(B)$.
The breakdown is that:
$$
x \in A \cap B \implies f(x) \in f(A \cap B)
$$
but not the other way around.  For example, $A = \{0,1\}$, $B = \{-1, 0\}$, $f(x) = x^2$.
In this case, $f(-1) = 1 \in f(A)$, but $-1 \notin A$.
On the other hand, if $f(x)$ is one-to-one, then
$$
f(b) = f(x) \implies b = x
$$
Also, by definition of $f(A)$, we have:
$$
    y \in f(A) \implies \exists \, a \in A \mid f(a) = y
$$
Substitute $f(x) = y$ and combine these, to get:
$$
f(x) \in f(A) \Leftrightarrow x \in A.
$$
Thus, if f is one-to-one, then the implication in the first line of your proof is bidirectional, and the proof works.
