Proof that $A \subset f^{-1}(f(A))$ 
Let $f: X \rightarrow Y$ be a function. $A \subset X$ and $B \subset Y$.
  Prove $A \subset f^{-1}(f(A))$.

Here is my approach. 
Let $x \in A$. Then there exists some $y \in f(A)$ such that $y = f(x)$. By the definition of inverse function, $f^{-1}(f(x)) = \{ x \in X$ such that $y = f(x) \}$. Thus $x \in f^{-1}(f(A)).$
Does this look OK, and how can I improve it?
 A: Your proof looks pretty good. The only thing to point out is when you said:

By the definition of inverse function, $f^{-1}(f(x)) = \{ x \in X$ such that $y = f(x) \}$. Thus $ x \in f^{-1}(f(A)).$

Two comments on this:


*

*This isn't usually called the inverse function -- we reserve that for when $f$ is invertible, and has a function $f^{-1}: Y \to X$. Instead, $f^{-1}$ here is called the inverse image, which is not a function $Y \to X$ but instead it takes subsets of $Y$ to subsets of $X$. It's confusing, I know, that they have the same symbol $f^{-1}$ for both.

*Your description of $f^{-1}(f(x))$ is a bit muddled, even though the reasoning is correct. It should be, $\{x' \in X \text{ such that } f(x') = f(x)\}$. That's the definition of $f^{-1}(a)$, where I put in $a = f(x)$. And I used $x'$ instead of $x$ because you don't want to mix up your two different $x$s.
A: A nice mnemonic on preimages is:$$x\in f^{-1}(C)\iff f(x)\in C\tag1$$
It is evident that: $$\forall x\left[x\in A\implies f\left(x\right)\in f\left(A\right)\right]$$
According to $(1)$ here $f(x)\in f(A)$ can be replaced by $x\in f^{-1}(f(A))$.
This results in:$$\forall x\left[x\in A\implies x\in f^{-1}(f(A))\right]$$
or equivalently: $$A\subseteq f^{-1}(f(A))$$
A: "there exists some $y\in f(A)$ such that $y=f(x)$" is a cumbersome way of expressing it.  I'd just say "let $y = f(x)$." Also, I would avoid  using the same letter, $x$, in two different senses, especially in view of the fact that not every point whose image under $f$ is equal to $f(x)$ needs to be the same as $x.$
A: Proposition. Let $X$ and $Y$ be sets. Let $f:X\to Y$. For each $A\in\mathscr{P}(X)$, $A\subseteq f^{-1}(f(A))$.
Proof. Let $A\in\mathscr{P}(X)$ be arbitrary.
\begin{align}
f^{-1}(f(A))&=\{z\in X:f(z)\in f(A)\}\\
&=\{z\in X:f(z)\in\{y\in Y:(\exists x\in A)[f(x)=y]\}\}\\
&=\{z\in X:(\exists x\in A)[f(x)=f(z)]\}.
\end{align}
$$A\subseteq f^{-1}(f(A)).$$
Remark. Let $X$ and $Y$ be sets. Let $f:X\to Y$. If $f$ is injective (i.e. one-to-one), then for each $A\in\mathscr{P}(X)$, $A= f^{-1}(f(A))$.
