Let $X$ and $Y$ be finite sets and $f: X \rightarrow Y$ be a function. Which one of the following statements is $\textbf{TRUE}$?

Let us analyze each statement for any function $f: X \to Y$:

1. Option A: $|f(A \cup B)| = |f(A)| + |f(B)|$ is false because $f(A)$ and $f(B)$ might not be disjoint, so $|f(A \cup B)| = |f(A) \cup f(B)| \le |f(A)| + |f(B)|$.

2. Option B: In general, $f(A \cap B) \subseteq f(A) \cap f(B)$, but equality does not necessarily hold unless $f$ is injective. For example, if $X = \{1, 2\}$, $Y = \{3\}$, $f(1)=f(2)=3$, $A = \{1\}$, and $B = \{2\}$, then $A \cap B = \emptyset \implies f(A \cap B) = \emptyset$, but $f(A) \cap f(B) = \{3\}$.

3. Option C: Using the same counterexample as in (B), $|f(A \cap B)| = 0$ while $\min\{|f(A)|, |f(B)|\} = 1$.

4. Option D: The preimage (inverse image) of an intersection preserves set operations perfectly:

   $$x \in f^{-1}(S \cap T) \iff f(x) \in S \cap T \iff (f(x) \in S \text{ and } f(x) \in T) \iff x \in f^{-1}(S) \cap f^{-1}(T)$$

   Hence, $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$ is always TRUE.

Thus, (D) is the correct option.