Doubt about the validity of a question's answers in MathSE - set of all sets with a given cardinality

Question

The question I am referring to is the following: Showing that the class of all sets of a particular cardinality is not a set.

OP presented his proof of the fact stated in the title by assuming the existence of such a set (which he calls $T_h$ where $h$ is the cardinal number in question) and by reconstructing Russel's Paradox, i.e. "the set $U$ of all sets does not exist". In doing so he simply says "let $a$ be any set" and the proceeds in showing that $a \in \cup T_h$ thus concluding that $\cup T_h = U$. However it seems to me that we are implicitly assuming that $U$ exists and then we are just proving that $T_h \subseteq U$.

I may be at fault so I am asking if OP's argument is correct or not, especially because no one else pointed this fallacy out under his question.

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As for the proof of the statement, I guess that we can reproduce Russel's argument:

Suppose $H:=\{X: |X|=h\}$ exists for some cardinal number $h$ (I shall use the definition of cardinal numbers as initial ordinal numbers). Define $R:=\{X \in \cup H: X \notin X\}$, then by Russel's argument $R$ can not belong to $\cup H$ (otherwise it would have to either belong or not belong to itself, which is impossible). However, we can always define a set from $h$ by replacing the least element with $R$ so that the new set $h_R$ has cardinality $h$. Therefore $h_R \in H$ and thus $R \in \cup H$, contradiction.

Answer 1

The argument shows that $\cup T_h = U$, not $T_h = U$. By the axiom of union, if $T_h$ were a set, then $U$ would be as well. But we know $U$ is not a set by the axiom of regularity. So $T_h$ is not a set.

Answer 2

You are right that the proof in that question fails to justify its claim that $\bigcup T_h$ is a set.

The union axiom says that if $\{T_h\}_h$ is a set, then $\cup \{T_h\}_h = \bigcup_h T_h$ is a set. But knowing that all the $T_h$ are sets doesn't tell us that $\{T_h\}$ is a set. (If it did, we could get to Russel's paradox just by saying every set $x$ is a set, so $\{x\}$ is the set of all sets!)

(Also, if the proof were valid, it would only disprove the claim that $T_h$ is a set for every cardinality $h$, showing that $T_h$ is not a set for at least one cardinality $h$. Your proof shows the stronger claim that for any cardinality $h$, $T_h$ is not a set.)