What is the flaw in what I just said?
Your argument correctly shows that the Goldbach conjecture cannot be false and undecidable. However, it can be true and undecidable.
This is equivalent to saying that it can be true in the standard model of, say, Peano arithmetic, but false in some nonstandard model: in a nonstandard model there may be a nonstandard counterexample to Goldbach's conjecture which is not an ordinary integer, and so which cannot be written down in the usual sense.
The reason it can be true and yet assuming it's false doesn't lead to a contradiction is that if it's undecidable, Peano arithmetic doesn't know it's true, so Peano arithmetic can't use the fact that it's true in a proof.
Edit- Added more thorough explanation
When they say the Riemann Hypothesis is true (or anyone says any mathematical assertion is true), they don't mean it is 'logically valid', they mean that it is true, i.e. that there are no nontrivial zeros of the zeta function.
It's not even clear without further elaboration what precisely it would mean for the Riemann Hypothesis to be logically valid. First we must cast the Riemann hypothesis (or some equivalent statement) as a sentence in some formal language, and only then we can start talking about interpretations and validity. Once we've done this, the idea of the sentence being "true" becomes a statement about one interpretation, usually called the 'standard' or 'intended' interpretation / model.
It is certainly not saying that it is valid in predicate logic, i.e. true in every interpretation, which is a very strong statement. More reasonable would be that it is a logical consequence of some effectively describable collection of axioms (perhaps PA if the language is arithmetic... more on that below), i.e. it is true in any model of those axioms. But even in this case there are generally many unintended models that don't have much bearing on ordinary mathematics, so they don't mean that either.
So the way to think about it is just like they say: if RH is false (in the standard interpretation), then its negation is provable, and hence it is not undecidable. So undecidable implies true (in the standard interpretation). If we want to be more model theoretic, we could say that if RH is false in the standard model, then it is false in all models, but really it's easiest to see that it is refutable directly, which implies by soundness that it is false in all models.
(And let me emphasize again, we've been pretty noncommittal about questions like 'what version of RH?', 'expressed in what language?', and 'model of what axioms?'... all of which needs to be fleshed out).
So the assertion in the video makes sense, and as they say, the reasoning is that if RH were false, it would be refutable. However, they gloss over the explanation, perhaps leaving you with the incorrect impression that it is trivial. Even the basic reasoning that if there's a counterexample (i.e. a nontrivial zero), we can check it, is annoyingly very close to being correct while still being very misleading.
To understand why this is an oversimplification, think about what would naively go into 'checking a zero'. The value of the zero itself may in principle contain an infinite amount of information (it's a complex number after all). There's no obvious reason why checking that it gives zero when plugged into the zeta function would be an effective procedure that translates to an effective proof. So its falsity is not "demonstrable by the existence of a non-trivial zero".
However, it has been known for some time that RH is equivalent to a $\Pi_1$ sentence of arithmetic, which is a sentence of the form "for all natural numbers $n$ the effectively computable property $P(n)$ holds" (and this fact isn't obvious unless you are a analytic number theorist well-versed in formalizing things in arithmetic). In this form it is obvious that it is refutable if false, since it is now clear that the procedure for verifying a given counterexample to $P(n)$ is effective, so any reasonable system will be able to turn this into a proof of $\exists n \lnot P(n)$
Original answer:
The truth they refer to is truth in the standard model, not validity. So the way to look at it is that if the Riemann Hypothesis is false in the standard model, then it is false in all models and hence not undecidable. The reason, as it says in the numberphile video, is that if it is false, then its negation is provable, and thus it is false in all models. This is due to the nontrivial theorem that the RH is equivalent to a $\Pi_1$ arithmetical sentence, so a (standard) counterexample to the Riemann Hypothesis will be witnessed by a finite arithmetical computation. (I haven’t watched the numberphile video (I’m traveling) but I’m guessing they oversimplified this... it would not be enough for a counterexample to exist... it needs to be provable.)
Best Answer
The issue here is how complicated is each statement, when formulated as a claim about the natural numbers (the Riemann hypothesis can be made into such statement).
For the purpose of this discussion we work in the natural numbers, with $+,\cdot$ and the successor function, and Peano Axioms will be our base theory; so by "true" and "false" we mean in the natural numbers and "provable" means from Peano Arithmetic.
We will say that a statement is "simple" if in order to verify it, you absolutely know that you do not have to check all the natural numbers. (The technical term here is "bounded" or "$\Delta_0$".)
For example, "There is a prime number small than $x$" is a simple statement, since verifying whether or not some $n$ is prime requires only checking its divisibility by numbers less than $n$. So we only need to check numbers below $x$ in order to verify this.
On the other hand, "There is a Fermat prime larger than $x$" is not a simple statement, since possibly this is false but only checking all the numbers above $x$ will tell us the truth of this statement.
The trick is that a simple statement is true if and only if it is provable. This requires some work, but it is not incredibly difficult to show. Alas, most interesting questions cannot be formulated as simple statements. Luckily for us, this "provable if and only if true" can be pushed a bit more.
We say that a statement is "relatively simple" if it has the form "There exists some value for $x$ such that plugging it in will turn the statement simple". (The technical term is $\Sigma_1$ statement.)
Looking back, the statement about the existence of a Fermat prime above $x$ is such statement. Because if $n>x$ is a Fermat prime, then the statement "$n$ is larger than $x$ and a Fermat prime" is now simple.
Using a neat little trick we can show that a relatively simple statement is also true if and only if it is provable.
And now comes the nice part. The Riemann hypothesis can be formulated as the negation of a relatively simple statement. So if the Riemann hypothesis was false, its negation was provable, so Riemann hypothesis would be refutable. This means that if you cannot disprove the Riemann hypothesis, it has to be true. The same can also be said on the Goldbach conjecture.
So both of these might turn to be independent, in the sense that they cannot be proved from Peano Arithmetic, but if you show that they are at least consistent, then you immediately obtain that they are true. And this would give us a proof of these statements from a stronger theory (e.g. set theory).
You could also ask the same about the twin prime conjecture. But this conjecture is no longer a relatively simple statement nor the negation of one. So the above does not hold for, and it is feasible that the conjecture is consistent, but false in the natural numbers.