An equivalence relation on $A$ has three properties: reflexivity, symmetry, and transitivity.
Reflexivity means that: "For all $a\in A$, $(a,a)\in R$." You have successfully added the pairs to make the relation reflexive by adding $(1,1)$, $(2,2)$, and $(3,3)$.
Symmetry means that: "If $(a,b)\in R$, then $(b,a)\in R$." You have successfully added the pairs (thus far) to make the relation symmetric by adding $(2,1)$ since $(1,2)$ is in the relation and $(3,2)$ since $(2,3)$ is in the relation.
Transitivity means that: "If $(a,b),(b,c)\in R$, then $(a,c)\in R$." You have not dealt with transitivity in your argument. In particular, there are pairs that you have in your relation of the form $(a,b)$ and $(b,c)$, but $(a,c)$ hasn't been added to the relation.
I teach these relations using dominos. In particular, for reflexivity, you've always got to have the dominos that have the same value on both sides. For symmetry, you're allowed to spin the dominos so that the left becomes the right. For transitivity, you're allowed to put two dominos together if they have the same value on one side (technically, this last description uses both symmetry and transitivity).
I don't want to give a partial answer because any partial answer is actually a full answer.