I understand how Cantor's diagonal argument can be used to prove that the real numbers are uncountable. But I should be able to use this same argument to prove two additional claims: (1) that there is no bijection $X \to \mathcal{P}(X)$ and (2) that there are arbitrarily large cardinal numbers.
I don't understand how to apply the diagonal argument, or a variation of its logic, in this case. I know how to prove Cantor's theorem. I suppose there exists a surjection $f: X \to P(X)$, consider the set $B = \{x \in X \mid x \not \in f(x)\}$, argue that there has to exist $y \in X$ such that $f(y) = b$, and then show that $y \in B$ and $y \not \in B$ both give contradictions. To show that there are arbitrarily large cardinals, I take $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$, $\mathcal{P}(\mathcal{P}(\mathbb{N}))$, and continue ad infinitum.
Is there a way that I'm silently using the diagonal argument here? Or is there a different way to think of these arguments?
Your solution to (2) is wrong. To show that there are arbitrarily large natural numbers divisible by $3$ we don't start with $0$, apply the successor function (or any increasing function), and hope for the best. We show that if $n$ is a number then one of $n+1,n+2,n+3$ is divisible by $3$ and all are strictly larger than $n$.
The same here. You're not supposed to iterate power sets from $\Bbb N$. How do you know that the sequence doesn't have an upper bound which is the largest cardinal? Instead, show that if $X$ is any set whatsoever, there is another set which has a strictly larger cardinality. For example, apply (1), but note that "there is no surjection from $X$ to $Y$" does not necessarily mean that $Y$ is larger, we also need to exhibit an injection from $X$ into $Y$.