The essence of your argument has nothing to do with well-orders. For example, we can also prove that for every real number $x$, the equation $x=x$ holds. That is, we proved $\forall x\in\mathbb{R}\ x=x$. Your suggestion is now to consider every particular real number $a$, and realize that we have a proof that $a=a$. Does this give uncountably many proofs, one for each real number? No, because $a=a$ is not a statement in the formal language, which you said is finite, and so $a=a$ doesn't even count as a statement in the formal language. If we were to expand the formal language with constant symbols to allow us to refer to $a$, then indeed, we could prove $a=a$, but in this case, the language itself would be uncountable and so the first part of your argument would break down.
Similarly, for the proper class case, we can prove $\forall x\ x=x$, where the universal quantifier ranges over all objects $x$, including every ordinal. Does this give a proper class of proofs, of the statements $a=a$ for every object $a$? No, again, because those statements are not in the formal language. If we augmented the formal language with constants for every object (or every ordinal), then indeed we would have a proper class number of proofs, but only because the language itself was already a proper class.
So ultimately, the solution to your confusion is to pay more attention to what exactly is in the formal language and what is not. If we can refer to the objects directly with constants in the formal language, then the claim that the language has a finite alphabet is false, invalidating the first part of the argument that there are only countably many proofs. And if we cannot refer to those objects with constants in the formal language, then we cannot have proofs of the statement $a=a$ or any other statement about $a$, since these assertions are not in the formal language.