I would like to try to prove that there is no set of all singletons without using powersets or relying on the theorem that there is no set which contains every set. Is it possible to complete this proof through proof by contradiction.
Show that there is no set to which every singleton (that is every set of the form $\{x\}$) belongs. Suggestion Show that from such a set we could construct a set to which every set belongs]
My attempt at the proof below:
Let $A$ be a set. We claim that $B$ is the set of singletons $B = \{ x \in A \mid x \in \{x\}\}$. It follows that $C = \bigcup B = \bigcup \{x\}$.
By definition of Union $x \in \bigcup \{x\}$ implies that there exists $x \in \{x\}$ therefore $\bigcup \{x\} \subseteq A$
By construction it follows that $x \in \bigcup \{x\}$ therefore $A \subseteq \bigcup \{x\}$ \ which implies that $A = \bigcup \{x\}$ then $B = \{x \in \bigcup \{x\} \mid x \in \{x\}\}$ but by definition of union there exists $x \in \{x\}$ such that $B = \{x \in \{x\} \mid x \in \{x\}\}$ which is a contradiction and so it follows $B = \{ x \in A \mid x \in \{x\}\}$ is not a valid set.
For any set $x$ it is true that $x\in\{x\}$. In particular, $x\in\{x\}$ for every $x\in A$, and therefore $B=A$. Since $A$ was specified to be a set, $B$ is a set; there is no contradiction here.
It appears that you went astray when you started to consider $\bigcup B$. This is simply $\bigcup A$, which means that for any $x$, $x\in\bigcup B$ if and only if there is an $a\in A$ such that $x\in a$.