Since you already know about the Alexandroff one-point compactification, let me begin by saying that the Stone-Cech compactification is at the other extreme, adding as many points at infinity as possible. To see what that could mean, let's consider some other compactifications of $\mathbb R$, starting with the most familiar, the extended real line, obtained by adjoining the two points $+\infty$ and $-\infty$ at the two ends of the line. Compared with the Alexandroff compactification, we're now distinguishing two different ways of "going to infinity". Some sequences that converged to $\infty$ in the Alexandroff compactification fail to converge in the extended real line because part of the sequence goes to the left and part to the right (e.g., $(-1)^nn$).
This idea can be extended, to produce "bigger" compactifications. If you visualize $\mathbb R$ as embedded in the plane as the graph of the sine function and then take its closure in the extended plane $(\mathbb R\cup\{+\infty,-\infty\})^2$, you get a compactification with a whole line segment at $+\infty$ (and another at $-\infty$). Similarly, embedding $\mathbb R$ in $3$-dimensional space as a helix, by $x\mapsto (x,\cos x,\sin x)$, we get a compactification with circles at the ends. Proceeding analogously with all bounded continuous functions $\mathbb R\to\mathbb R$ (in place of $\cos$ and $\sin$), in a very high-dimensional space (in fact, $2^{\aleph_0}$ dimensions), you get one of the standard constructions of the Stone-Cech compactification of $\mathbb R$. Roughly speaking, it separates, into different points at infinity, all of the possible "ways to go to $\infty$" in $\mathbb R$.
The analogous story works for discrete spaces $X$ in place of $\mathbb R$ (except that I don't need to say "continuous" because all functions on a discrete space are continuous). A point in the Stone-Cech remainder of a discrete space $X$ should be thought of as a "way to go to $\infty$" in $X$. But how can such "ways" be described?
Well, in any compactification, each point $p$ at infinity is in the closure of the original space $X$, and so we can describe its location relative to $X$ by the trace on $X$ of its neighborhood filter, i.e. by $\mathcal F=\{U\cap X: p\in\text{interior}(U)\}$ (where $U$ refers to subsets of the compactification). Note that $\mathcal F$ can't contain any finite subsets of $X$, because such subsets are closed in the compactification (as in any $T_1$ space) and thus disjoint from suitable neighborhoods of any point $p$ at infinity.
For the Stone-Cech compactification of a discrete space $X$, this filter $\mathcal F$ must have one additional property, namely that we cannot have two disjoint subsets $A,B$ of $X$ both meeting all the sets in $\mathcal F$. The reason is that then "going to infinity" in $A$ and in $B$ would be two different ways to go to infinity, both leading to the same point $p$.
This additional property of the filter $\mathcal F$ is equivalent to saying that $\mathcal F$ is an ultrafilter. So this is how ultrafilters enter the picture of Stone-Cech compactifications of discrete spaces.
One then ordinarily continues by saying that, (1) since the filter $\mathcal F$ associated with any $p$ at infinity is an ultrafilter, and different $p$'s must correspond to different ultrafilters (in order for the compactification to be Hausdorff), we might as well identify the points $p$ with the ultrafilters $\mathcal F$, and (2) for the sake of uniformity, we might as well identify the principal ultrafilters (which haven't been used yet) with the points of $X$. With these conventions, one can prove (using compactness) that every ultrafilter on $X$ gets identified with a point in the Stone-Cech compactification of $X$. So the Stone-Cech compactification of a discrete space can be identified with the set of ultrafilters on $X$. Finally, one should verify that the topology is necessarily the one you described.
(The following answer is mostly repeated from my comments above, but I have added the observation that the least $\alpha$ such that $\mathcal{G}_\alpha$ is an ultrafilter can (consistently) be a successor ordinal. I'm working in ZFC.)
There seems to be a problem with the exercise...
First, there is an easy observation to see that the $\mathcal{G}_\alpha$'s, for $\alpha<2^{\aleph_0}$, are not all of cardinality $<2^{\aleph_0}$. Recall $\mathcal{G}_0$ is the Frechet filter, so $\mathcal{G}_0$ is countable.
Let $\beta$ be least such that $A_\beta$ is infinite, coinfinite (i.e. infinite and $\omega\backslash A_\beta$ is infinite). Note then that $\mathcal{G}_\beta=\mathcal{G}_0$, and $A_\beta\in\mathcal{G}_{\beta+1}$. Note that for every
$X\subseteq\omega\backslash A_\beta$, we have $A_\beta\subseteq A_\beta\cup X\in\mathcal{G}_{\beta+1}$. But there are $2^{\aleph_0}$-many such sets $X$,
hence $2^{\aleph_0}$-many such $A_\beta\cup X$, so $\mathcal{G}_{\beta+1}$ has cardinality $2^{\aleph_0}$.
So the exercise is incorrect. What if we weaken the exercise, and just demand
that $\mathcal{G}_\alpha$ fail to be an ultrafilter for $\alpha<2^{\aleph_0}$?
Then, consistently, it is still incorrect...
A base for an ultrafilter $U$ on $\omega$
is a set $B\subseteq U$ such that for every $X\in U$ there is $Y\in B$ with $Y\subseteq X$. Let $\mathfrak{u}$ be the least cardinal $\kappa$ such that for some nonprincipal ultrafilter $U$ on $\omega$, there is a base $B$ of $U$ of size $\kappa$. (This is well-defined, since there are non-principal ultrafilters on $\omega$.) Trivially then $\mathfrak{u}\leq 2^{\aleph_0}$. A direct combinatorial argument shows $\aleph_1\leq\mathfrak{u}$. So assuming CH, we have $\mathfrak{u}=2^{\aleph_0}=\aleph_1$, and the statement you wanted to prove can be proved using this extra assumption.
But Kunen constructed a model satisfying $2^{\aleph_0}=\aleph_{\omega_1}$ and $\mathfrak{u}=\aleph_1$; see Kunen "Set theory: An introduction to independence proofs" Chapter 8, Exercise A10. In that exercise, one constructs a model with an ultrafilter on $\omega$ with a base of size strictly smaller than the continuum.
Also, Baumgartner and Laver showed that, assuming ZFC is consistent, so is ZFC + $2^{\aleph_0}=\aleph_2$ + $\mathfrak{u}=\aleph_1$. (See their paper "Iterated perfect set forcing".) (These points are mentioned in section 9 of Blass's chapter in the Handbook of set theory, "Combinatorial cardinal characteristics of the continuum", p. 448. That section also contains much more related information.)
So it could be that $\mathfrak{u}<2^{\aleph_0}$. Assume this, and let $U$ be a non-principal ultrafilter with a base $B$ of cardinality $\mathfrak{u}$. Let $\left<A_\alpha\right>_{\alpha<\mathfrak{u}}$ enumerate $B$. Let $\left<A_\alpha\right>_{\mathfrak{u}\leq\alpha<2^{\aleph_0}}$ then be such that $\left<A_\alpha\right>_{\alpha<2^{\aleph_0}}$ enumerates $\mathcal{P}(\omega)$. Define the sequence of filters $\mathcal{G}_\alpha$ as you do in the question. Note that since $U$ is a filter, we get by induction that $\mathcal{G}_\alpha\subseteq U$ and $A_\alpha\in\mathcal{G}_{\alpha+1}$, for each $\alpha<\mathfrak{u}$. But then $B\subseteq\mathcal{G}_{\mathfrak{u}}$, and since $B$ is a base for $U$, therefore $\mathcal{G}_{\mathfrak{u}}=U$, which is an ultrafilter. But $\mathfrak{u}<2^{\aleph_0}$, contradicting the weaker version of the exercise.
Note that in the construction above, the $\mathcal{G}_\alpha$ are not ultrafilters
for $\alpha<\mathfrak{u}$, just by definition of $\mathfrak{u}$.
So $\mathfrak{u}$ is the least $\alpha$ such that $\mathcal{G}_\alpha$ is an ultrafilter there.
Finally, let me just observe that, again assuming that $\mathfrak{u}<2^{\aleph_0}$,
it can also be that $\mathcal{G}_\alpha$ is first an ultrafilter at some successor $\alpha$. For this, let $A\subseteq\omega$
be infinite, coinfinite, let $U_1$ be a nonprincipal ultrafilter on $A$,
having a base of cardinality $\mathfrak{u}$,
and let $U_2$ be a nonprincipal ultrafilter on $\omega\backslash A$,
also with such a base. (Note that if $U$ is any nonprincipal ultrafilter
on $\omega$ with a base of cardinality $\mathfrak{u}$, then we can easily
produce such ultrafilters on $A$ or on $\omega\backslash A$, by shifting
then according to a bijection $\pi:\omega\to A$ or $\pi:\omega\to\omega\backslash A$.)
Let $F$ be the filter on $\omega$ of sets of the form $C\cup D$ where $C\in U_1$ and $D\in U_2$ (note it is a filter). Then $F$ is not an ultrafilter on $\omega$, since $A\notin F$ and $\omega\backslash A\notin F$. But note that $U_1$ is a base for a unique ultrafilter on $\omega$ (as is $U_2$). Let $B_1=\{A_{1\alpha}\}_{\alpha<\mathfrak{u}}$ be a base for $U_1$ and $B_2=\{A_{2\alpha}\}_{\alpha<\mathfrak{u}}$ be a base for $U_2$. Let $A_\alpha=A_{1\alpha}\cup A_{2\alpha}$. Let $A_{\mathfrak{u}}=A$. Then let $\left<A_\alpha\right>_{\mathfrak{u}<\alpha<2^{\aleph_0}}$ enumerate the rest of $\mathcal{P}(\omega)$. Suppose we use this sequence and define the $\mathcal{G}_\alpha$'s. Then it is straightforward to see that $\mathcal{G}_{\mathfrak{u}}=F$, which is not an ultrafilter, but since $A_\mathfrak{u}=A$, that $\mathcal{G}_{\mathfrak{u}+1}$ is the ultrafilter with base $U_1$. So $\mathfrak{u}+1$ is the least stage where we get an ultrafilter in this case.
Best Answer
The simplest property that I can think of (right now) that provably (in ZFC) distinguishes some non-principal ultrafilters on $\mathbb N$ from others is "weak P-point", which means "not in the closure in $\beta\mathbb N$ of a countable set of other non-principal ultrafilters." The existence of weak P-points is a theorem of Kunen; the existence of non-principal ultrafilters that are not weak P-points is trivial (take any countably infinite set of non-principal ultrafilters and take any other point in their closure).