EDT. Clarification: $f:A\rightarrow B$
EDT. Changed the first line in the proof according to responses / comments.
Proof.
Let $a \in f^{-1}(S)$. Then, $f(a)\in S$.
Because $S\subset T$, if $f(a)\in S$ then also $f(a)\in T$ and therefore $a \in f^{-1}(T)$.
For all $a \in f^{-1}(S)$ we then have $a \in f^{-1}(T)$ and therefore $f^{-1}(S)\subset f^{-1}(T)$
QED
Your start is again wrong. You have to take $a\in f^{-1}(S)$ then you can say $f(a)\in S$ (and not vice versa).
Every time you want to prove $M\subseteq N$, you have to take $x\in M$ (and just in $M$, no $f(M)$ or $f^{-1}(M)$ or something else, just $M$)! Then you prove that $x$ is also in $N$.