I haven't had a chance to watch the video, but the argument you've described makes an important error which is incidental to this question; I'll treat this error below.
There's a subtlety here, which crops up all the time in logic or whenever we're doing logic-y things: what system are we working in?
Whenever you use words like "provable," "undecidable," or similar, you're always hiding an important bit of context; provability only makes sense with respect to some system of axioms. Goedel's incompleteness theorem is a great example of this: given an appropriate theory $T$, we find a sentence $p$ and prove "$p$ is true but $T$ doesn't prove $p$" - but we prove this in a theory stronger than $T$ itself, so this isn't a contradiction.
The exact same thing is going on here. We have some appropriate (in particular, sufficiently strong) theory $T$, and some stronger theory $S$. Suppose $S$ proves that $T$ doesn't disprove RH; there are two relevant facts kicking around here:
If $T$ doesn't disprove RH, then RH is true. (This is provable in $T$ itself.)
$T$ doesn't disprove RH. (This is provable in $S$, but not necessarily in $T$.)
In order to conclude "RH is true," we need both of these facts; $S$ has them, but $T$ doesn't necessarily. So we can think of this as a kind of meta-reasoning - we're using two different kinds of provability here ($S$-provability and $T$-provability), and our argument only gives us a proof of RH in the first sense.
In fact, we can prove (again, in a stronger theory than $T$ :P) that $T$ doesn't prove that it doesn't disprove RH (whew, that's a mouthful!). This is because if $T$ is inconsistent, then it disproves everything; put another way, if $T$ doesn't disprove something, then $T$ is consistent. $T$ can prove this if $T$ is sufficiently powerful, so if $T$ proved "I don't disprove RH" then $T$ would prove "I am consistent," and this can't happen as long as $T$ is consistent by Goedel's second incompleteness theorem. So the theory $T+Con(T)$, which is stronger than $T$, proves "$T$ doesn't prove that $T$ doesn't disprove RH" (and of course we can replace RH with anything here).
OK, now the error:
Suppose there was a counterexample to RH. This counterexample is some complex number $z$. But there are uncountably many complex numbers, and only countably many expressions in the language of our theory; so most complex numbers can't be defined in the language we're working in! This means that for most complex numbers $z$, our theory $T$ can't even express the statement "$z$ is a counterexample to RH," let alone prove it. And now we have a problem: what if there are counterexamples to RH, but they're all undefinable?
So the argument there is really missing something important. This can, however, be fixed: it turns out that RH is equivalent to an appropriate statement about natural numbers, such that our theory $T$ can verify a counterexample to it if one exists.
But this is a really important point, and it's annoying to see it glossed over!
