Disproving the statement A×B⊆C×D if and only if A⊆C and B⊆D

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I know that a counterexample can disprove this statement I tried a lot to disprove it without using a counterexample If A⊆C and B⊆D let (x,y)∈ (A×B) ⇒x∈A and y∈B ⇒x∈C and y∈D(from the assumption) ⇒A×B⊆C×D Similarly, let (A×B)⊆(C×D) let (x,y)∈(A×B) ⇒ (x,y)∈(C×D)(from the assumption) i.e., x∈A and y∈B ⇒ x∈C and y∈D ⇒ A⊆C and B⊆D But this proves the statement instead of disproving it I have just begun discrete maths, and I can't figure out where I am going wrong. It would be extremely helpful if someone could tell me where I am going wrong?