Prove that $\operatorname{tr.deg}(\mathbb{C/Q})=\mathfrak{c}$ (where $\mathfrak{c}$ is the cardinality of $\mathbb{R}$) using the continuum hypothesis.
$Proposition$ If $E/F$ is an algebraic extension then $|E|=\aleph_0 |F|$
Proof:
Let $S$ be a trancendental basis of the extension $\mathbb{C/Q}$.Then $|S| \leqslant \mathfrak{c}$.Suppose that $|S|< \mathfrak{c}$. Then, by the continuum hypothesis, $S$ is finite or countable.
Let $S=\{s_1,s_2, \ldots \}$ a countably infinite set.
Then let $A_n=\mathbb{Q}(s_1,,,s_n)$
$(\dagger)$ We have that every $A_n$ is a countable set and $\bigcup_{n=1}^{\infty}A_n=\mathbb{Q}(S)$ which is countable as a countable union of countable sets.
Also by definition of the trancendental basis we have that $\mathbb{C}/\mathbb{Q}(S)$ is an algebraic extension thus by the above proposition $|\mathbb{C}|=\aleph_0 |\mathbb{Q}(S)|=\aleph_0 \aleph_0=\aleph_0$ - deriving a contradiction.
Is this proof correct?
I'm not sure about if the equation of the sets in $(\dagger)$ is correct.
Yes, the proof you suggest is correct. To see why, note that $\mathbb{Q}(S)$ is obtained by taking the rational functions in $|S|$ variables. If $S$ is finite, then there are only countably many such rational functions.
If you know basic cardinal arithmetic, you can in fact show that $|\mathbb Q(S)|=|S|+\aleph_0$, and therefore the continuum hypothesis is not needed for the proof.