f is a function from $X \rightarrow Y$
Let's consider options one by one and also consider whether we can have a failure case for each of the options
(A)
For any subsets A and B of X,$|f(A∪B)|=|f(A)|+|f(B)|$
This case can fail if A and B have atleast one common element, in that case AUB will count only total number of distinct elements in A and B and then it will find image of all such elements and give the count
But the RHS expression will separately give count for the number of images of A+Number of images of B.
Assume A={A,B} and B={B,C}. $|f(A)|=2$ and $|f(B)=2|$ and $|f(A \cup B)|=3$
So this option is clearly wrong.
(B)
For any subsets A and B of X,$f(A∩B)=f(A)∩f(B)$
This can fail in case of Many to one function where A and B are completely disjoint but the image set of A and B overlap.
Take A={A,B} and B={C,D} LHS will come to be empty but RHS will have 2.
So this option also ruled out.
And through this reasoning also I rule out option (C)
(D)
For any subsets S and T of Y, $f^{−1}(S∩T)=f^{−1}(S)∩f^{−1}(T)$
If they have defined inverse of the function f , this means f is invertible and hence one-to-one and onto.
Since f is bijective, whenever any subsets of Y, S and T overlap, their corresponding images in $f^{-1}$ would also overlap.
And if they don't overlap, then their images also won't.
ANS-D