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?

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.
