I have interpreted the exercise in the following way: We are working in ZFC. In particular, this means that any element of a set is also a set. I have looked on the relation $\in$ for some simple sets which can be obtained from the empty set $\emptyset$.
When I look at the table of contents of Devlin's Joy of Sets (Fundamentals of Contemporary Set Theory), I saw that the author deals with axiomatic set theory only in the second chapter of this book. So it is possible that I have somehow misinterpreted what he wanted in this exercise.
Let us start by two trivial observations. If $x=\emptyset$, then all properties are vacuously true. If the set $x$ has one element, then $R$ is connected and antisymmetric, since we never have $a\ne b$.
The relation $\in$ on any set $x$ must be antisymmetric, since $a\in b\in a$ contradicts Axiom of regularity.
If $x=\{\emptyset,\{\emptyset\}\}$ then the relation $\in$ is not symmetric, just take $a=\emptyset$ and $b=\{\emptyset\}$. Notice that relation $\in$ on this set is connected and it is also transitive.
Let $x=\{\emptyset,\{\{\emptyset\}\}\}$. If we denote $a=\emptyset$ and $b=\{\{\emptyset\}\}$, then neither $a\in b$, nor $b\in a$. So the relation $\in$ on this set is not connected. However, as it is the empty relation, it is transitive.
Let $x=\{a,b,c\}$ where $a=\emptyset$, $b=\{\emptyset\}$ and $c=\{\{\emptyset\}\}$. Then $a\in b$ and $b\in c$, but $a\notin c$, so the relation $\in$ on this set is not transitive. It is also not connected, since neither $a\in c$ nor $c\in a$.
$A \ \mathrm{x}\ A=\{(a,b) | a=a_1...a_9 , b=b_1...b_9, a,b \in A\}$
i) Reflexive: is $(a,a) \in R$, where $a=a_1...a_9$? Yes, because $a_1=a_1$.
ii) Transitive: if $(a,b) \in R$ and $(b,c) \in R$ do we have $(a,c) \in R$? No, take for example $a_1=b_1$, $a_2 \neq b_2$ and $b_1 \neq c_1$, $b_2 = c_2$. We have $a_1 \neq c_1$ and $a_2 \neq c_2$, thus $(a,c) \notin R$.
iii) Symmetric: if $(a,b) \in R$ it means that $a_1=b_1$ or $a_2=b_2$. Therefore we have $b_1=a_1$ or $b_2=a_2$ i.e. $(b,a) \in R$.
iv) Antisymmetric: if $(a,b) \in R$ and $(b,a) \in R$ then we can have $a \neq b$: take for example $a_3 \neq b_3$. Thus it's not antisymmetric
v) It can't be an equivalence relation since it's not transitive.
Best Answer
If $SR$ is symmetric then $xSRy\iff ySRx$.
We have to show $xSRy\iff xRSy$.
So it suffices to show $xRSy\iff ySRx$. So we need to show:
$\exists z(xRzSy)\iff\exists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $\Box$