Please Comment on the validity of presented proof of the given theorem.
Theorem. Given that $R$ is a relation on $A$, and $S$ is the transitive closure of $R$, If $R$ is symmetric, then so is $S$.
Proof. Assume that $R$ is a relation on $A$, $S$ is the transitive closure of $R$ and $R$ is symmetric.
Let $\mathcal{F} = \{T\subseteq A\times A|R\subseteq T \land\forall x\in A\forall y\in A\forall z\in A(xRy\land yRz\implies xRz)\}$
Now assume that $(x,y)\in R$ where $x$ and $y$ are arbitrary members of $A$ as $R$ is symmetric it follows that $(y,x)\in R$ and $S$ is the transitive closure of $R$ then it must be that $R\subseteq S$ and so $(y,x)\in S$ which implies that $(x,y)\in S^{-1}$ since $x$ and $y$ were arbitrary it follows that $R\subseteq S^{-1}$.
Now Assume that $(z,y)\in S^{-1}$ and $(y,x)\in S^{-1}$ where $z,y$ and $x$ are arbitrary members of the set $A$, from this it is evident that $(x,y)\in S$ and that $(y,z)$ but since $S$ is transitive it follows that $(x,z)\in S$ which entails that $(z,x)\in S^{-1}$. since $x$ and $y$ were arbitrary it follows that $S^{-1}$ is transitive.
We have now established that $R\subseteq S^{-1}$ and that $S^{-1}$ is transitive but then $S\in \mathcal{F}$ and since $S$ is the transitive closure of $R$ and thus satisfies $\forall T\in\mathcal{F}(S\subseteq T)$ consequently $S\subseteq S^{-1}$.
Now Assume that $(x,y)\in S$ where as before $x$ and $y$ are arbitrary in $A$ it follows that $(x,y)\in S^{-1}$ and thus $(y,x)\in S$ generalizing our argument to any ordered pair in $S$ establishes the symmetry of $S$.
Best Answer
If $aSb$, then there is some $x_1,.. x_j$ with
$. . aRx_1, x_1Rx_2,.. x_jRb$.
As $R$ symmetric,
$. . . bRx_j, x_jRx_{j-1},.. x_1Ra$;
so $bSa$.