Proof: if $R$ is symmetric then so is $R^{-1}$ This is one problem I have been solving in Velleman's How to prove book:

Suppose $R$ is a relation on $A$, prove that if $R$ is symmetric, then
  so is $R^{-1}$.

Now this is my proof:

Suppose $R$ is symmetric. Let $a$ and $b$ be an arbitrary element in
  $A$ such that $(a,b) \in R^{-1}$. From $(a,b) \in R^{-1}$ it follows
  that $(b,a) \in R$. Since $R$ is symmetric, it follows that $(a,b) \in
 R$. So, $(b,a) \in R^{-1}$. Therefore, if $(a,b) \in R^{-1}$ then
  $(b,a) \in  R^{-1}$. Since $a$  and $b$ are arbitrary, it follows that
  $R^{-1}$ is  symmetric

.
I have been verifying my proof from here and they have given a slightly different proof:

Suppose R is symmetric. Let x,y be arbitrary elements of A s.t.
  (x,y)∈R. It follows from our assumption that (y,x)∈R also. Then $(x,y) \in R^{-1}$ and $(y,x) \in R^{-1}$ as well. Since x, y are arbitrary,
   so If R is symmetric then so is $R^{-1}$.

In the above proof they are staring from the assumption that $(x,y) \in R$ which I think is not correct. Am I missing something here?

Both your proof and the one you quote are correct. There's no reason why we couldn't start with $(x,y)\in R$.
However, the proof could be made clearer by saying:
$$(x,y)\in R\Longrightarrow (y,x)\in R^{-1}\quad\text{(definition)}$$
$$(x,y)\in R\Longrightarrow (y,x)\in R\quad\text{(symmetry)}$$
$$(y,x)\in R\Longrightarrow (x,y)\in R^{-1}\quad\text{(definition)}$$
instead of:

It follows from our assumption that $(y,x)\in R$ also. Then $(x,y)\in R^{-1}$ and $(y,x)\in R^{-1}$ as well.

That way you see exactly where everything is coming from - but that's just my personal opinion.
