The original question I had was vaguely "Does consistency of ZFC decide Continuum Hypothesis?", but I'm not even sure whether it is a valid question. So let me ask this question in prior to that question.
By Gödel's second incompleteness, for every consistent axiomatic system $\mathsf{A}$, $\mathsf{Con}(\mathsf{A})$ is an independent statement.
That said, let $\mathsf{ZFC}_0$ be a synonym of $\mathsf{ZFC}$. Since we hope that $\mathsf{ZFC}_0$ is consistent, I shall just take $\mathsf{Con}(\mathsf{ZFC}_0)$ as an axiom, and call the resulting system $\mathsf{ZFC}_1$.
Yet $\mathsf{ZFC}_1$ is just another system, and is prone to the incompleteness. Don't worry, I shall just take $\mathsf{Con}(\mathsf{ZFC}_1)$ as an axiom, and call the resulting system $\mathsf{ZFC}_2$. ad infinitum.
Now I have $\mathsf{ZFC}_n$ for every finite ordinal $n$. Yet I won't stop here. Why wouldn't I take $\mathsf{Con}(\mathsf{ZFC}_n)$ for all $n$? Since these axioms are countable, the resulting system is valid (By "valid" I mean "describable in a formal language"). I shall call it $\mathsf{ZFC}_\omega$.
I still won't stop here. By transfinite induction, it follows that $\mathsf{ZFC}_{\omega+1}$, $\mathsf{ZFC}_{\omega \cdot 2}$, $\mathsf{ZFC}_{\omega^2}$, $\mathsf{ZFC}_{\omega^\omega}$, $\mathsf{ZFC}_{\epsilon_0}$, $\mathsf{ZFC}_{\epsilon_1}$, $\mathsf{ZFC}_{\zeta_0}$, $\mathsf{ZFC}_{\Gamma_0}$ and so on are still valid, since they all have countably many axioms. (Though I'm quite unsure about $\mathsf{ZFC}_{\omega_1^{CK}}$)
Now I got a problem. Can I take $\mathsf{Con}(\mathsf{ZFC}_\alpha)$ for all countable ordinal $\alpha$? The resulting system, $\mathsf{ZFC}_{\Omega}$, would have uncountably many axioms, so the transfinite induction cannot confirm that it is valid. Must I stop here and call it a day?
That said, the original question actually was "Is there an ordinal $\alpha$ where $\mathsf{ZFC}_\alpha$ decides Continuum Hypothesis?", but that should be another post.