Proving that $A\subseteq f^{-1}(f(A))$ and that if $f$ is injective $A=f^{-1}(f(A))$.

this is an intro to analysis problem on Set Theory, Relations and Functions:

Let $f:S\to T$ be a function and $A\subseteq S$. Prove that $A\subseteq f^{-1}(f(A))$ and that $A=f^{-1}(f(A))$ if $f$ is injective.

Proof: (1) Let $x\in A$, then $\{x\}\subseteq A$, then $f(\{x\})\subseteq f(A)$, then $f^{-1}(f(\{x\}))\subseteq f^{-1}(f(A))$. Since $\{x\}\subseteq f^{-1}(f(\{x\}))$, then $\{x\}\subseteq f^{-1}((f(A))$, i.e. $x\in f^{-1}(f(A))$, thus $A\subseteq f^{-1}(f(A))$.

(2) Now let $x\in f^{-1}(f(A))$, since $f$ is injective, there is a distinct $x\in S$ such that $f(x)=y\in T$, then $f^{-1}(f(x))=f^{-1}(y)=x$, then $f^{-1}(f(x))\in f^{-1}(f(A))$, i.e. $x\in A$, thus $f^{-1}(f(A))\subseteq A$. Since $A\subseteq f^{-1}(f(A))$ and $f^{-1}(f(A))\subseteq A$, $A=f^{-1}(f(A))$.

Any comments would be of great help! Thanks!

Your argument in (1) is correct but much more complicated than necessary. Just note that $f^{-1}\big[f[A]\big]=\{x\in S:f(x)\in f[A]$ by the definition of $f^{-1}$. If $x\in A$, then certainly $f(x)\in f[A]$ and hence $x\in f^{-1}\big[f[A]\big]$, so $A\subseteq f^{-1}\big[f[A]\big]$.

Your argument in (2) is a bit confused, though I think that you have the right idea. You already know that $A\subseteq f^{-1}\big[f[A]\big]$, so you want to show that if $x\in f^{-1}\big[f[A]\big]$, then $x\in A$. Suppose that $x\in f^{-1}\big[f[A]\big]$; then by definition $f(x)\in f[A]$, so there is an $a\in A$ such that $f(x)=f(a)$. But $f$ is injective, so $x=a\in A$, and $A=f^{-1}\big[f[A]\big]$.

Your proof seems correct, but complicated.

If $f\colon S\to T$ is a function and $X\subseteq T$, then, by definition, it is equivalent to say, for $s\in S$,

$$ s\in f^{-1}(X) $$ and $$ f(s)\in X $$

If $s\in A$ (for $A\subseteq S$), you certainly have $f(s)\in f(A)$, hence $s\in f^{-1}(f(A))$.

If $A\ne f^{-1}(f(A))$, then there is $x\in f^{-1}(f(A))$, $x\notin A$. But $f(x)\in f(A)$, so there is $y\in A$ with $f(x)=f(y)$. Since certainly $x\ne y$, you have proved that $f$ is not injective.