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.
Good question!
If you were to consider the system PA + $\neg$G, wouldn't G still hold in the sense that you could never find a counterexample?
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.
Best Answer
I think it's helpful to consider $(i)$ a simpler statement than Goodstein's theorem and $(ii)$ a weaker theory than $\mathsf{PA}$. The reason for the former is that Goodstein's theorem has lots of complexity which isn't really germane to the general logical issues at play; the reason for the latter is that there is a particular technical result (Tennenbaum's theorem) which prevents us from having any too-nicely-describable nonstandard models of $\mathsf{PA}$ at all, and again that makes the logical issues look more troubling than they really are.
So let's take as our statement "Every number is either even or odd" which we can formally express as $$(\star):\quad\forall x\exists y(y+y=x\vee y+y+1=x),$$ and let's take as our theory Robinson's arithmetic $\mathsf{Q}$. This theory, in contrast with $\mathsf{PA}$, has lots of easily-describable nonstandard models. In particular, my favorite nonstandard model of $\mathsf{Q}$ is the set $\mathfrak{P}$ of polynomials in a single variable $x$ with integer coefficients and nonnegative leading coefficient, ordered by eventual domination as $x\rightarrow\infty$. (More snappily, take the ring $\mathbb{Z}[x]$, order it so that $p>q$ whenever $p$ is a polynomial of higher degree than $q$, and look at the nonnegative elements of that ordered ring.)
Now it's easy to show that $\mathfrak{P}$ does not satisfy $(\star)$: just consider the polynomial $x$ by itself. To us, this is not a natural number, but "within $\mathfrak{P}$" it looks perfectly fine. Perhaps belaboring the point, there is a unique way embedding of $\mathbb{N}$ inside $\mathfrak{P}$, and $x$ is not in the image of that embedding.
The case of $\mathsf{PA}$ and Goodstein's theorem is fundamentally no different: we can find a (necessarily nonstandard) model $\mathfrak{M}$ of $\mathsf{PA}$ such that $\mathfrak{M}$ "thinks" that Goodstein's theorem fails. But any of the things $\mathfrak{M}$ thinks are counterexamples to Goodstein will be nonstandard, just as $x$ was a nonstandard element of $\mathfrak{P}$ in the example above.
The following old answers of mine cover related material: on $\mathfrak{P}$ specifically and on nonstandard models in general (the latter is a bit technical unfortunately).