Attempt at proving the class of all cardinals is a proper class

Question

Define $C=\{\alpha:\alpha=|x|$ for some set $x$$\}$ as the class of all cardinals. ($|x|$ being the cardinality of the set $x$)

It will be enough to prove $C$ is a proper class by showing $On\subseteq C$

Since we can take it to be an established fact that $On$ is a proper class.

Let $\alpha \in On$. Assume $\alpha \notin C$

So $\neg\exists x \in V $(the universal class) such that $|x|=\alpha \implies$

So $\neg\exists x \in V $ st. $\alpha$ is the least ordinal such that $\alpha \approx x$

But it's trivially true that $\alpha \approx \alpha \in On \subseteq V$. So $\alpha \in V$. This would be a contradiction since we assumed there is no such set.

Therefore, $\alpha \in C \implies On\subseteq C$ and then $C$ is a proper class.

$\textbf{Edit:}$ As pointed out by Pedro, the initial assumption of $On\subseteq C$ is false. In fact, $C \subseteq On$ is true - since all cardinals are necessarily ordinals.

Answer 1

It turns out it would be enough to prove $On \subseteq \bigcup C$:

Suppose $C$ is a set. Then we know $\bigcup C$ is also a set from the unions axiom. If the above claim holds true then it would follow that $On$ is a set $-$ which is known not to be true.

Let $\alpha \in On$

A previously proven theorem I will assume is Cantor's Theorem:

For any set $x$, we know $x\prec p(x)$

We know $\alpha$ is a set since all ordinals are sets.

It follows $\alpha \prec p(\alpha)$

Every set is bijectively equivalent to its own cardinal so $p(\alpha)\approx |p(\alpha)|\in On$

The least ordinal that satisfies the property $p(\alpha)\approx \beta$ is the cardinal of $p(\alpha)$, we know at least one such ordinal exists by the Numeration Principle.

So, $\alpha \prec p(\alpha)\approx|p(\alpha)|\in On $

It has also been shown previously if $\alpha \prec p(\alpha)$ $\implies |\alpha| \lt |p(\alpha)|$, but $|\alpha|\approx \alpha$

So $\alpha \lt |p(\alpha)|\implies \alpha \in |p(\alpha)|$ since $\alpha, |p(\alpha)|\in On $

Now, $ p(\alpha)$ is a set by the power set axiom. So we can deduce from the definition of $C$ that $| p(\alpha)|\in C\implies \alpha \in \bigcup C$

So, it holds true that $On\subseteq \bigcup C$, and we have the necessary contradiction.