The transcendence degree of $\mathbb{C}$ over $\mathbb{Q}$ is exactly the cardinality of the continuum. Certainly it can't be countable, since any field with countable transcendence degree over a countable field is countable. On the other hand, it has cardinality at most that of the continuum. So we're already done assuming CH. But I think the result holds even without CH; can anyone confirm / deny?

The transcendence degree of either $\mathbb{C}$ or $\mathbb{Q}_p$ over $\mathbb{Q}$ is exactly the cardinality of the continuum. Certainly it can't be countable, since any field with countable transcendence degree over a countable field is countable. On the other hand, either transcendence degree is at most that of the continuum. So we're already done assuming CH. But I think the result holds even without CH; can anyone confirm / deny? (Edit: Yes, as the other answer shows.)