Show that for any arbitrary relation $R (D \cap E) = R(D) \cap R(E)$

Question

So I'm supposed to prove that given any arbitrary relation $R \subseteq A \times B$ And given two subsets of $A$, $D$ and $E$, the following equality holds $R (D \cap E) = R(D) \cap R(E)$. I was able to prove that $R (D \cap E) \subseteq R(D) \cap R(E)$, but I'm pretty sure that the subset relationship doesn't go the other way around, in fact I think I found a counterexample. Consider the sets $A = \{a, b\}$ and $B = \{1\}$, then consider the relation $R \subseteq A \times B$ given by $R = \{(a,1), (b,1)\}$. Lastly consider the subsets of $A$, $D = \{a\}$ and $E = \{b\}$. Then if I'm not mistaken: $R (D \cap E) = R(\emptyset) = \emptyset$ and $R(D) \cap R(E) = \{1\} \cap \{1\} = \{1\}$ So am I correct on my observation that the subset relation only goes one way but not the other? Is there an error on the book I'm reading? Thanks for your help!