How to prove ($i_A \cap R = \emptyset \implies R \subseteq A \times A $ - irreflexive) by contrapositive

Question

$i_A = ${$ (a,a)| \forall a \in A $} is an identity relation on a set $A$.

I am having a hard time proving this statement by contrapositive. What I mean by "proving by contrapositive" is at first I began with direct proof, so I assumed ($i_A \cap R = \emptyset$) is true and then tried to prove by contrapositive ($R \subseteq A \times A$ - irreflexive) i.e. ($ \forall a : a \in A \implies (a,a) \notin R$) is true, so to prove ($\forall a : a \in A \implies (a,a) \notin R$) is true by contrapositive I have to prove ($\forall a: (a,a) \in R \implies a \notin A$) is true. Essentially, I have to prove for an arbitrary $a$ that if $(a,a)$ is an element of $R \subseteq A \times A$ then $a$ is not an element of $A$, but what confuses me is that if $(a,a)$ is an element of $R \subseteq A \times A$ then $a$ has to be an element of $A$ because R is a subset of $A \times A$. Did I make any mistake in the process? Is it possible to prove this statement like this? If it can be done, what is the method?

Just for the record, I managed to prove this statement by combining direct proof and proof by contradiction, as well as only using direct proof.