Can there be an $\aleph_2$ or even greater? Can there be an $\aleph_2$ or even greater? I was reading about Cantor’s Diagonal Proof and learned about $\aleph_1$ and $\aleph_0$. Would it make sense for there to be an $\aleph_2$ by just using the same method that Cantor used but on $\aleph_1$?
 A: The answer to the headline question is yes. There is an $\aleph_1$ an $\aleph_2,$ an $\aleph_3,$ an $\aleph_4,$ etc. There is even an $\aleph_\omega$ where $\omega$ is the same thing as $\aleph_0$ (just usually denoted $\omega$ instead to emphasize that it's natural to think of it as an ordinal here rather than a cardinal). And then there's the next cardinal larger than than that: $\aleph_{\omega+1}.$ And so on. For any ordinal $\alpha$ there is a cardinal $\aleph_{\alpha.}$
The relationship between $\aleph_1$ and $2^{\aleph_0}$ - which is the cardinal that Cantor's proof shows is strictly larger than $\aleph_0$ - is not as clear as your question suggests. Because $\aleph_1$ is defined to be the very next largest size than $\aleph_0,$ we have $2^{\aleph_0}\ge \aleph_1.$ But they are not necessarily equal... the conjecture that they are equal is called the Continuum Hypothesis and is one of the most legendary problems in all of mathematics. It has been shown to be not provable one way or the other on the basis of the standard axioms of set theory (as well a slew of plausible axioms that go beyond the standard ones). Opinions over the years on the size of $2^{\aleph_0}$ have ranged from "it's $\aleph_2$" to "it's immensely large, well larger than $\aleph_\omega$" to "it's $\aleph_1$" (this is the CH and was Cantor's opinion, though few set theorists seem to believe it nowadays) to "it is not a well-defined problem."
(Edit: As J.G. remarked in the comments, the “very large” option is consistent with the axioms of set theory. In fact there is an unboundedly large class of alephs to which the axioms cannot decide if $2^{\aleph_0}$ is equal. There are cardinals that we can prove not equal to $2^{\aleph_0};$ as JG mentions, $\aleph_\omega$ is the smallest such. In fact there is a simple condition (countable cofinality) that determines whether a cardinal is provably not $2^{\aleph_0}$ or if it is undecidable.)
The fact that we can indeed iterate Cantor's argument like you suggest (it really shows that for any set $X,$ its power set $\mathcal P(X)$ is strictly larger) does mean that we can always find a larger set, so does guarantee the existence of the alephs (though see Andres's answer about the role of the axiom of choice in all of this... I am assuming it freely), even if the sets you get by iterating don't necessarily correspond to successive alephs. There is a name for these sets, actually: the beths. They are defined as $\beth_1 = 2^{\aleph_0},$ $\beth_2 = 2^{\beth_1},$ etc. 
A: Yes, there is an $\aleph_2$ and an $\aleph_3$, and there are alephs beyond all the $\aleph_n$. 
A suitable version of Cantor's diagonal proof is perfectly general and shows that, for any set $X$, $|X|<|\mathcal P(X)|$. In particular, if $X$ has size $\aleph_1$, then $\mathcal P(X)$, the power set of $X$, has size strictly larger and therefore admits a subset of size $\aleph_2$.
The appropriate way to define $\aleph_1,\aleph_2,\dots$, however, is to first consider the ordinals, and then define the subclass of so-called initial ordinals (these are the alephs). One can prove, using just properties of ordinals, that given any ordinal $\kappa$ there is an ordinal of strictly larger size. This argument, in particular, does not require any use of the axiom of choice. 
On the other hand, to show that if $|X|=\aleph_\alpha$ then $\mathcal P(X)$ admits a subset of size $\aleph_{\alpha+1}$ makes use of this axiom. For instance, it is consistent with set theory without the axiom of choice that there is no subset of the reals (or, equivalently, of $\mathcal P(\mathbb N)$) of size $\aleph_1$. This is perhaps a bit confusing, but without the axiom of choice, many infinite cardinals are incomparable, and there may be cardinals for which it makes no sense to talk of the (or even a) smallest larger cardinal. 
A: Since the formulation of your question seems to be based on confusing $2^{\aleph_0}$ with $\aleph_1$, and other answers are talking about $\aleph_0, \aleph_1, \aleph_2, \dots$, I'll answer the other interpretation of the question: Is there a set larger than $2^{\aleph_0}$, and can it be constructed with the same method.

The answer is yes, and actually quite easily. For a set $S$, consider the set of all subsets of $S$, called the power set of $S$ and denoted by $2^S$. Now Cantor's diagonal argument in most general sense says that for each $S$, the set $2^S$ always has strictly larger cardinality than $S$.
In particular, $\aleph_0$ is the cardinality of $\mathbb{N}$, and $2^{\aleph_0}$ is the cardinality of $2^\mathbb{N}$ (the set of sets of natural numbers). So we know that $2^{\aleph_0} > \aleph_0$. Now if you consider the set $2^\mathbb{N}$ and take the power set of that, $2^{2^\mathbb{N}}$ (the set whose elements are all sets whose elements are subsets of $\mathbb{N}$), you have cardinality $2^{2^{\aleph_0}}$ which again is strictly larger than $2^{\aleph_0}$.
Or, more easily: $2^{\aleph_0}$ also happens to be the cardinality of $\mathbb{R}$ (the set of real numbers). So if you take the power set of $\mathbb{R}$, you get a set with cardinality $2^{2^{\aleph_0}}$.
And you can continue taking power sets to get even larger and larger sets.
