Clarifying the definitions of image and pre-image

If $f(x) \in C$, it is indeed true that it does not follow that $x \in C$. Think about the function $\mathbb{R} \ni x \mapsto x^2\in \mathbb{R}$. Even though $f(\sqrt{2}) = 2 \in \mathbb{Q}$, obviously $\sqrt{2} \notin \mathbb{Q}$. Why should the image of an element being in a set mean that this element itself is in the set? Functions can be very weird and send sets to all kinds of other sets. This is where the reciprocal implication fails in $(1)$.

On the other hand, $f(x) \in D$ does indeed imply that $x \in f^{-1}(D)$. This is because we define $f^{-1}(D)$ to be the set of all elements whose images are in $D$, that is: $x \in f^{-1}(D) \iff f(x) \in D$. So that's clearly not the problem here. It's not that $x \in f^{-1}(D) \implies f(x) \in f(f^{-1}(D))$ is false either, that's clearly true (because $a \in A$ clearly implies $f(a) \in f(A)$ always). All this can only mean one thing: your proof for $(2)$ is wrong. Because if it was right, we would get the reciprocal implication for free. The problem here is the same one as in $(1)$: $f(a) \in f(B)$ does not imply $a \in B$.