I was looking over the proofs we use to show that $\text{CH}$ is independent to $\text{ZFC}$. The methods used have to be higher than $\text{ZFC}$, otherwise the proof could be formalized in $\text{ZFC}$ and would show that $\text{ZFC + CH}\vdash \text{Cons(ZFC + CH)}$, thus showing that $\text{ZFC + CH}$ is inconsistent by Gödel's incompleteness theorem.
I was wondering, if we are able to use such proofs for the independence of these formulas (and we accept them as true, despite them being above finitary reasoning), could we ever be able to show that $\text{ZFC}$ is consistent using the same level or reasoning? Would we accept this as a strong argument? Do any mathematicians object to the independence proofs because they can't be finitary? Do we care about proving $\text{ZFC}$ consistent or do we just assume it is?
The consistency proofs are indeed "$\sf\operatorname{Con}(ZFC)\to\operatorname{Con}(ZFC+CH)\land\operatorname{Con}(ZFC+\lnot CH)$", and while we can in fact formalise these proofs in $\sf ZFC$ itself, and indeed it is far more convenient for us to do that, we can actually formalise them in weak theories like $\sf PRA$.
Now, your question is ultimately kind of meaningless, since we are not proving $\sf ZFC\vdash\operatorname{Con}(ZFC+CH)$, but rather an implication. But it is true that $\sf ZFC$ proves that every of its finite fragments is consistent. You can conclude from this that if $\sf ZFC$ is indeed consistent, then any model of $\sf ZFC+\lnot\operatorname{Con}(ZFC)$ must contain non-standard integers, since no "real axioms" can lead to a proof of contradiction which has a "really finite length" using "real inference rules" (in other words, any proof that can be coded by a real finite integer cannot be a proof that $0=1$ from $\sf ZFC$).
You can use this as a philosophical argument in favour of $\sf ZFC$, but of course there is a circularity to it. It only works if you assume that $\sf ZFC$ is already consistent, otherwise the whole point is moot. Inconsistent theories prove everything anyway.