Here's the flaw in the compactness argument you sketched. In the model you construct, the interpretation of $c$ will indeed be a nonstandard number whose Goodstein sequence does not halt in a standard number of steps. But Goodstein's theorem has a universal quantifier for the number of steps, so nonstandard "Goodstein sequences" are also permitted. That is, when the theorem says "there is a finite sequence", in a particular model that "finite" sequence might have the length of a nonstandard number, which cannot be excluded by the sequence of sentences in your argument.
One way to find an extension of PA where Goodstein's theorem is provable is to add enough transfinite induction to PA to formalize the usual proof of Goodstein's theorem. The details of how to do that are not so bad, although they may take too much time for an elementary class.
There is a lot to address here. You might be getting at any of the following points:
The existence of $\Pi_1^0$ statements: A statement is called $\Pi_1^0$ if, if it were false, it could be disproven by a finite amount of data. Goldbach's conjecture, "every even number is the sum of two primes", is $\Pi_1^0$, because to disprove it you just have to give an even number $2n$, and a nontrivial factor of $2n-p$ for each prime $p < 2n$. A $\Pi_1^0$ statement, if undecidable, is always true. See Wikipedia for the rigorous definition and much more.
It is far from the case that all interesting problems in number theory are $\Pi_1^0$. Consider the twin prime conjecture, "there are infinitely many pairs of primes which differ by $2$." This could be undecidable in two different ways. Maybe it is true, but we can't prove it. Or, maybe there are no twin primes past $10^{10^{10}}$, and we can't prove that. Either way, no finite amount of data will settle the issue.
Goodstein's theorem is not $\Pi_1^0$: Contrary to your statement, a finite amount of data cannot disprove Goodstein's theorem. Suppose that you knew that the Goodstein sequence starting at $100$ did not terminate. How could you convince me o this with any finite computation?
Now, in fact, Goodstein's theorem is true, because induction up to $\epsilon_0$ is valid. See this MO question for more. But consider the Collatz conjecture. Like Goodstein's theorem, it says that a recursively defined sequence starting at any integer eventually terminates. It is perfectly possible for Collatz to be either undecidable and true or undecidable and false.
ZFC does prove more than PA: The proof of Goodstein's theorem can be formalized in ZFC. So can the intuitive proof that PA is consistent, namely, "of course it's consistent, it is a set of true statements about $\mathbb{Z}$!" So it is true that being undecidable in ZFC is a lot stronger than being so in PA.
A point of philosophy: Some mathematicians will defend the belief that all statements about $\mathbb{Z}$ are either true or false (though they may be unprovable from the current axioms) but this need not be true about sets. The idea here is that there is only one set of integers, but there are many equally good versions of set theory. Scott Aaronson defends this view here; JDH defends the "more than one set theory" here. (I don't know whether or not JDH thinks there is more than one version of $\mathbb{Z}$.)
Note that this is much stronger than the claim that all $\Pi_1^0$ statements are true or false. Scott, for example, presumably believes that the twin prime conjecture is either true or false, even though no finite amount of computation could ever settle it.
I have not thought enough about this question to form an opinion; it is ultimately a matter of philosophy, not math.
Best Answer
Good question!
This gets at an important subtlety here - the issue of models. Any first-order theory like PA, PA+$\neg$G, ZFC, ... has (assuming it's consistent!) many models. Some theories, like PA, have an "intended model," but by the compactness theorem every (interesting) theory will have lots of nonisomorphic models. In particular, in addition to the standard model $(\mathbb{N}; +, \times, 0,1, <)$ of PA, there will also be lots of "nonstandard" models of PA. These models are difficult to describe (for good reason), but very vaguely they're ordered semirings (like $\mathbb{N}$) with "$\mathbb{N}$-ish" properties - in particular, they have a definable notion of exponentiation which behaves the way we're used to and lets the model talk about "finite" sequences of "numbers." The key difference is that a nonstandard model of PA will have elements which are actually infinite.
Such a nonstandard element can constitute an apparent failure of G within that model. If $M$ is a model of PA+$\neg$G, then there is some $m\in M$ which according to $M$ is a counterexample to $G$. However, this $m$ won't actually be a "true" natural number: it won't be $0$, or $1$, or $1+1$, or ...
At this point it might help to get a little more concrete. As I said above, nonstandard models of PA are difficult to visualize. However, if we look at a weaker theory of arithmetic - say, Robinson's Q - things get much better. Q is a very weak theory indeed, and has many easily-describable nonstandard models. Perhaps the nicest of these is what I'll call $\mathcal{P}$, the set of integer-coefficient polynomials in one variable $x$ with positive leading coefficient (okay fine and also the zero polynomial should be included).
$\mathcal{P}$ has a number of bizarre arithmetic properties. For instance, "every number is even or odd" is false in $\mathcal{P}$! The polynomial "$x$" is neither even nor odd, since no element $p$ of $\mathcal{P}$ satisfies either $p+p=x$ or $p+p+1=x$. There are many other examples of such strangeness. (Incidentally, this shows that you need induction to prove that every number is either even or odd! :P)
So what's going on? Well, of course in "reality" (that is, $\mathbb{N}$) every number is even or odd. The problem is that $\mathcal{P}$, while satisfying the axioms of $Q$, has "too many numbers" including some very strange ones which as far as $\mathcal{P}$ is concerned are in fact counterexamples to the statement "every number is either even or odd." So this shows that the statement "every number is either even or odd" can't be proved from $Q$ alone.
Exactly the same thing is going on with Goodstein's theorem G; it's just that the theory in question being PA, the "bad" models are much harder to visualize and so the independence is more mysterious.