Suppose $f: X \rightarrow Y$, and is one-to-one, and let $A \subseteq X$, prove that $f^{-1}[f[A]] = A$. Suppose $f: X \rightarrow Y$, and is one-to-one, and let $A \subseteq X$, prove that $f^{-1}[f[A]] = A$.
EDIT: Actually, this identity should hold even if $f$ is not one-to-one (injective), right?
This is completely intuitive and logical, but I can't think of a proof strategy to use to prove it? I'm familiar with induction and element chasing, but they don't seem to work here.
Here is the beginning of my element chasing proof to prove that $f^{-1}[f[A]] \subseteq A$.
Start with arbitrary element: $a \in f^{-1}[f[A]]$ \
Apply definition of preimage: $a \in \{x \in X | f(x) \in f[A] \}$ \
Simplify and convert to boolean expression: $a \in X \land f(a) \in f[A]$ \
Apply definition of image: $a \in X \land f(a) \in \{y \in Y | y = f(x), x \in A \}$ \
I'm stuck here...
 A: 
"Apply definition of image: $a \in X \land f(a) \in \{y \in Y \mid y = f(x), x \in A \}$".

The set should be written as $\{y\in Y\colon \exists x\in A(y=f(x))\}$, as you've written, it doesn't make sense.
You wish to conclude that $a\in A$.
Consider $f(a)$. One has $f(a)\in \{y\in Y\colon \exists x\in A(y=f(x))\}=\{f(x)\colon x\in A\}$, so there exists $x\in A$ such that $f(x)=f(a)$.

"Actually, this identity should hold even if $f$ is not one-to-one (injective), right?"

Can you now answer your own question?
The other inclusion holds universally (i.e., it doesn't require $f$'s injectivity).
A: If $f$ is not injective, it is not true.
Example: $X=\{0,1\},Y=\{0\},A=\{0\}$ and $f=$ the only possibility; then $f^{-1}(f(A))=X$.
In the general case, the only you can say is $f^{-1}(f(A))\supseteq A$: $f^{-1}(f(A))=\{x\in X \mid f(x)\in f(A)\}$ and this clearly contains $A$.
Assume $f$ is injective.
Take $x\in f^{-1}(f(A))$; by definition, $f(x)\in f(A)$, so there must exist an $a\in A$ such that $f(x)=f(a)$; by injectivity $x=a\in A$ and you get the reverse inclusion.
