I need to proof that the identity relation $id_A : A \rightarrow A$ is the only relation on $A$ which is both a function and an equivalence relation. I have already proven that $id_A$ is a function and an equivalence relation. Yet I need to prove the uniqueness. My approach is the following. Let there be an arbitrary relation $R$ which is a function and an equivalence relation, then $R=id_A$.
Proof: Suppose $R$ is not $id_A$, but a relation of the form $(a,b) \in A \times A$ with $a,b \in A$. Symmetry demands that for every pair $(a,b) \in A$ there exists a pair $(b,a) \in A$. The relation is a function so there does not exist a pair $(a,c) \in A$ with $b \neq c$. Yet transitivity needs to hold, so $(a,b) \wedge (b,c) \Rightarrow (a,c)$. Yet $(a,c) \in A$ does not exist. So a relation of a different form than $(a,a)$ cannot exist therefore $R=id_A$.
Is this proof correct, make any sense?
You only need to use the reflexive property. Clearly, $1_A$ is an equivalence relation. On the other hand, if $R$ is an equivalence relation that contains an off-diagonal element $(x,y)$, so that $x\neq y$ then as $(x,x)\in R$ we have $\left \{ (x,x),(x,y) \right \}\subset R$ so by definition, $R$ can not be a function.