What if there's an element that is in the $S \cap T$ that also maps to $y$? In other words, what if $f$ isn't injective? I honestly stopped reading there.
Usually when showing that a set is a subset of another, you want to take a direct approach. I'm not saying that a contradiction wouldn't work, and I believe that in more complicated situations it would, but for simple properties like this, you should try a direct approach first:
I'll be informal (or at least not use symbols), so you can seen if you understand the proof and write it yourself.
You pick an element in the set on the left, and you see that its preimage sits in $S\cap T$. This just follows from the definition of the preimage. So the preimage sits in both $S$ and $T$. Now apply $f$ to the elements in the preimage, what do you get?