Similar proof to the class of all sets is a proper class Assume Russells paradox: The collection $R=\{x\mid x\not\in x\}$ does not define a set
and the axiom of subsets: Let $\Phi(x)$ be a definite well defined proerty, let $x$ be any set, then $\{y\in x\mid\Phi(y)\}$ is a set.
I know that letting $V$ denote the class of all sets. Then $V$ is a proper class.
Proof: If $V$ was a set, then we should have that $R=\{x\in V\mid x\not\in x\}$ by the axiom of subsets defines a set. Contradiction! Therefore $V$ does not define a set.
Question: Prove that the class of all singleton sets (i.e. those of the from $\{x\}$) is not a set
Answer: Let $V$ denote the class of all singleton sets. If $V$ was a set, then we should have that $R=\{\{x\}\in V\mid \{x\}\not\in \{x\}\}$ by the axiom of subsets defines a set. Contradiction! Therefore $V$ does not define a set.
Is this a sufficient answer? Is there more to say given that we are now considering the class of all singleton sets? Do I need to define a function mapping a set $x$ to the singleton set $\{x\}$ for the answer to be sufficient?
 A: I would go about it a different way. 
Let $Z$ be the class of singleton sets. 
Suppose $Z$ is a set. 
Then by the Axiom of Unions $\bigcup Z$ is a set also.
So $\bigcup Z \in V$, as $V$ is the class of all sets. 
But notice that $\bigcup Z = \{x|(\exists \{x\} \in Z) s.t. (x \in \{x\})\}$
So it seems that $\bigcup Z = V$. 
Just to make sure this is the case,
Clearly then $\bigcup Z \subseteq V$ 
If $b \in V$, then $\{b\} \in Z$ and so $b \in \bigcup Z$
So $V \subseteq Z$ and thus $\bigcup Z = V$ 
Since $V$ is a proper class, $\bigcup Z$ cannot be a set.
From this we conclude that $Z$ is not a set.
A: No, your answer is not sufficient.
The Russell paradox, which you have used to show that the class of all sets is a proper class, essentially says that given a set $A$, there is a subclass $B$ of $A$ which is not an element of $A$.
In the case of "the class of all sets", if it were a set, and a subclass of a set is a set, then the class defined by the Russell paradox is a set, but cannot be a member of $V$, which is a contradiction since $V$ is the class of all sets.
To use this for the case of the class of all singletons, you must produce something that a priori has the chance to be an element of the class. In this case, a singleton. But $\{\{x\}\mid \{x\}\notin\{x\}\}$ is not a singleton, so there is no reason to expect that it will be an element of the class of all singletons.
Instead, suppose that $V_1=\{\{x\}\mid x=x\}$ is the class of singletons. If $V_1$ is a set, what can you say about $\bigcup V_1$?
(You can also find injections from the class of all sets into $V_1$, and a surjection from $V_1$ onto the class of all sets; if your set theory is capable of showing that these conditions suffice for showing that $V_1$ is a proper class, this is another reasonable way to prove something is a class.)
A: The reason that in Russell's pardox $R=\{x\mid x\not\in x\}$ does not define a set is that $$R \in R \iff R \notin R.$$
To translate that into singletons would be $R' = \big\{\{x\} \in V \mid \{x\} \notin x\big\}$ and then $$\{R'\} \in R' \iff \{R'\} \notin R'.$$
I hope that helps $\ddot\smile$
