So... after some input in comments it seems you are considering a function $f:X\to Y$ and that your aim is to show that $X=f^{-1}(f(X))$. You say you already know that $f^{-1}(f(X))\subseteq X$, hence you are interested in the other inclusion $X\subseteq f^{-1}(f(X))$. Let us prove the more general statement that, for every $Z\subseteq X$, one has $Z\subseteq f^{-1}(f(Z))$. Here we go.
Let $z\in Z$. One wants to show that $z\in f^{-1}(f(Z))$.
By definition, $f^{-1}(f(Z))=\{x\in X\mid\exists y\in f(Z),\,f(x)=y\}$, so one wants to exhibit some $y\in Y$ such that $y\in f(Z)$ and $f(z)=y$, right? Well, consider $y=f(z)$... Then $y\in f(Z)$ by construction (because $z\in Z$) and $f(z)=y$ hence $z$ fits the definition of $f^{-1}(f(Z))$, that is, $z\in f^{-1}(f(Z))$.
This proves that $Z\subseteq f^{-1}(f(Z))$ for every $Z\subseteq X$, in particular $X\subseteq f^{-1}(f(X))$ hence you are done.
Note that, as mentioned by others, in general for $Z\subset X$, the other inclusion $f^{-1}(f(Z))\subseteq Z$ may be false (and actually, it is false for some $Z\subset X$ as soon as $f$ is not injective).