confusion of proving f(A)∩f(B) a subset of f(A∩B)? f:A↦B and S⊆A,T⊆A, then f(S)∩f(T)⊆f(S∩T)

We know that it is not.By setting A:{1,-3} B:{1,3},f(x)=$x^2$.

But what if f is one-to-one function?

How can we prove that ∀y(y∈(f(S)∧f(T)))⇒∃x(x∈S∧x∈T))?
Since you are talking about $f(A) $ and $f(B) $ we can assume the function is onto also. Then you see the inverse exists and inverse images are well behaved with $\cup$ and $\cap$. And by bijectivity onto the image $f^{-1} (f(A)) =A$. 
In general if you don't want to use the above argument you can say the function has left inverse and using that you can achieve the same result. 

So $f^{-1} (f(A) \cap f(B)) = f^{-1}(f(A)) \cap f^{-1}(f(B)) =A\cap B $ applying $f$ both the sides you will get the result.
Since you are talking about $f(A) $ and $f(B) $ we can assume the function is onto also. Then you see the inverse exists and inverse images are well behaved with $\cup$ and $\cap$. And by bijectivity onto the image $f^{-1} (f(A)) =A$. 
In general if you don't want to use the above argument you can say the function has left inverse and using that you can achieve the same result. 

So $f^{-1} (f(A) \cap f(B)) = f^{-1}(f(A)) \cap f^{-1}(f(B)) =A\cap B $ applying $f$ both the sides you will get the result.