Proving that $C$ is a subset of $f^{-1}[f(C)]$

Since you want to show that $C\subseteq f^{-1}\big[f[C]\big]$, yes, you should start with an arbitrary $x\in C$ and try to show that $x\in f^{-1}\big[f[C]\big]$. You cannot reasonably hope to show that $f^{-1}\big[f[\{x\}]\big]=x$, however: there’s no reason to think that $f$ is $1$-$1$, so there may be many points in $A$ that $f$ sends to the place it sends $x$.

Let $x\in C$ be arbitrary. For convenience let $E=f[C]\subseteq B$. Now what elements of $A$ belong to the set $f^{-1}\big[f[C]\big]=f^{-1}[E]$? By definition $f^{-1}[E]=\{a\in A:f(a)\in E\}$. Is it true that $f(x)\in E$? If so, $x\in f^{-1}[E]=f^{-1}\big[f[C]\big]$, and you’ll have shown that $C\subseteq f^{-1}\big[f[C]\big]$.

Copied from my answer to I am issues with proving the following problem: $f^{-1}(f(A)) ⊃ A$, which was closed as a duplicate of this question just before I posted it.

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$$ f^{-1}\left(f(A)\right)=\left\{x:f(x)\in f(A)\right\}\tag{1} $$ Note that if $x\in A$, then $f(x)\in f(A)$, and by $(1)$, $x\in f^{-1}\left(f(A)\right)$. Therefore, by definition, we have $$ A\subset f^{-1}\left(f(A)\right)\tag{2} $$ However, if $f$ is not injective, then $f^{-1}\left(f(A)\right)$ may indeed contain elements not present in $A$; for example let $f:\mathbb{Z}\mapsto\mathbb{Z}$ be defined by $$ f(x)=\left\lfloor\frac x2\right\rfloor\tag{3} $$ and let $A$ be the set of even integers. Then $f(A)=\mathbb{Z}$ and $$ A\subsetneq\mathbb{Z}=f^{-1}\left(f(A)\right)\tag{4} $$