My professor provided this proof however, I am not convince that she used a valid argument. Since after assuming there exist an arbitrary mapping Φ that should show an isomorphism between D1 and D2, she then defined that mapping Φ... which I think defeats the purpose of Φ being arbitrary.
Suppose D1 is isomorphic to D2 then ∃Φ s.t. if (a,b)∈A(D1) then (Φ(a),Φ(b))∈A(D2). If we let Φ(1)=a, since (1,2)∈A(D1) then (a,Φ(2))∈A(D2). This implies that Φ(2)=b. Also, (2,3)∈A(D1) then (b,Φ(3))εA(D2) which means Φ(3)=c. Notice that (3,2)εA(D1) but (Φ(3),Φ(2))∉A(D2). Therefore, due to a contradiction, D1 is not isomorphic to D2. Q.E.D.
At this point, I do understand what the proof is saying but I think that if we are going to define Φ as such then we also need to exhaust all the other possible mappings since Φ is no longer arbitrary. Or is there something that I am missing here? We know there exist a far more easier proof but she wants us to only use the formal definition of isomorphism. 
The only assumption she arbitrarily made about $\phi$ is that $\phi(1)=a$.
In the case $\phi(1)\neq a$, that should also be checked (you're right about that), you reach a contradiction by noticing that $\exists x \in D2$ s.t. $(x,\phi(1))\in A(D2)$, but $\not\exists y \in D1$ s.t. $(y,1)\in A(D1)$ : you cannot find an antecedent to this $x$.