Questions regarding Cantors' Theorem The proof of Cantor's Theorem in the Wikipedia Article goes like this:

Two sets are equinumerous (have the same cardinality) if and only if there exists a one-to-one correspondence between them. To establish Cantor's theorem it is enough to show that, for any given set $A$, no function $f$ from $A$ into the power set of $A$, can be surjective, i.e. to show the existence of at least one subset of $A$ that is not an element of the image of $A$ under $f$. Such a subset is given by the following construction: $$ B = \{ x \in A : x \notin f(x) \}. $$
This means, by definition, that for all $x$ in $A$, $x ∈ B$ if and only if $x ∉ f(x)$. For all $x$ the sets $B$ and $f(x)$ cannot be the same because $B$ was constructed from elements of $A$ whose images (under $f$) did not include themselves. More specifically, consider any $x ∈ A$, then either $x ∈ f(x)$ or $x ∉ f(x)$. In the former case, $f(x)$ cannot equal $B$ because $x ∈ f(x)$ by assumption and $x ∉ B$ by the construction of $B$. In the latter case, $f(x)$ cannot equal $B$ because $x ∉ f(x)$ by assumption and $x ∈ B$ by the construction of $B$.
Thus there is no $x$ such that $f(x) = B$; in other words, $B$ is not in the image of $f$. Because $B$ is in the power set of $A$, the power set of $A$ has a greater cardinality than $A$ itself.

First, I don't quite understand this construction.
It keeps asking whether or not $B$ and $f(x)$ are equal, but $B$ a set of elements of $A$, and the image of $f(x)$ (EDIT: "... the image of f") is a set of sets (elements of the power set, mapped to from elements of $A$) - one is a set of elements and one is a set of sets, of course they are not ever going to be equal...?
Also, how can $B$ be in the power set of $A$? Isn't the power set of $A$ all of the possible subsets of $A$, and isn't B a set of elements of $A$?
Second, isn't there a simpler way of proving this?
What is wrong with this proof that the cardinality of the power set is strictly greater than the cardinality of the set:

Let $A$ be the set $\{ a_{1}, a_{2}, a_{3}, ... \}$ and $f$ be this function from $A$ to $P(A)$: $f(a_{n})$ = $\{ a_{n} \}$. Then the image of $A$ under $f$ is one-to-one (injective) because every element $a_{n}$ has a corresponding element in $P(A)$, but it is not onto (surjective) because there are many more elements of the Power Set (i.e. for every $a_{n}$, there are other elements of $P(A)$ than $\{ a_{n} \}$ containing $a_{n}$). Every element of $A$ is accounted for in the domain of $f$, but there are many elements in the image of $f$ unaccounted for. Therefore there cannot be a bijection (one-to-one and onto function) from $A$ to $P(A)$, and $P(A)$ has a strictly greater cardinality than $A$.

Is this not sufficient to show that $A$ has to few elements for there to be a bijection from $A$ to $P(A)$?
 A: First of all, $A$ might not be countable. $A$ is just an arbitrary set. So it isn't necessarily enumerated by the integers; in fact Cantor's theorem holds without even appealing to the axiom of choice, so it should hold for sets which cannot be enumerated by any ordinal anyway.
Secondly, $B$ as given is a subset of $A$. And $f$ is a function taking $x\in A$ and $f(x)\in\mathcal P(A)$, meaning $f(x)$ is a subset of $A$. So there's no issue with comparing them as "types" (both are sets anyway).
Finally, the point in the proof is that if $f$ is surjective, then the set $B$ as defined is in its range, namely there is some $x\in A$ such that $f(x)=B$. Then either $x\in B$ and we have a contradiction or $x\notin B$ and we have a contradiction.
On a side note, your "proof" really just appeals to the general Cantor's theorem, since it says something of the form "Well, the power set has more elements, so it has uncountably many elements!", but how do you know that it has uncountably many elements? Exactly by proving Cantor's theorem. So you can't quite make that argument in your proof.
A: QUOTE "$B$ a set of elements of $A$, and the image of $f(x)$ is a set of sets (elements of the power set, mapped to from elements of $A$) - one is a set of elements and one is a set of sets, of course they are not ever going to be equal...? END QUOTE
You refer to "the image of $f(x)$" rather than "the image of $f$".  In this context, that is a serious mistake, guaranteed to lead to incomprehension.  For any member $x$ of $A$, the set $f(x)$ is a subset of $A$ and thus a member of the power set of $A$.  The image of $f$ (not of $f(x)$) is a set of subsets of $A$, not a subset of $A$.
QUOTE Also, how can $B$ be in the power set of $A$? Isn't the power set of $A$ all of the possible subsets of $A$, and isn't $B$ a set of elements of $A$? END QUOTE
To say $B$ is "in" the power set of $A$ means $B$ is a member of the power set of $A$.
That means $B$ is a subset of $A$.
That is the same as saying $B$ is a set of elements of $A$.
QUOTE Second, isn't there a simpler way of proving this? END QUOTE
I doubt it.  This is one of the simplest mathematical proofs imaginable.  It seems as if your confusion is really about the meaning terminology rather than about the way the proof is organized.
QUOTE Then the image of $A$ under $f$ is one-to-one (injective) because END QUOTE
That's wrong.  The image of $A$ under $f$ cannot be one-to-one because that image is not a function.  I suspect what you mean is that $f$ is one-to-one.
QUOTE is one-to-one (injective) because every element $a_n$ has a corresponding element in $P(A)$ END QUOTE
That is not what one-to-one means.  That is in fact true of functions that are not one-to-one.  For example $x\mapsto x^2$ on the real line is not one-to-one, but it is still the case that every member $x$ of the real line has a corresponding square.  You should review the definition of one-to-one.  "One-to-one" means that two different elements of the domain never have the same image.  That is indeed true of this function, but not for the reason given.
You have correctly proved that this one function is not surjective.  But that is not what was to be proved.  What was to be proved is that no function with this domain and this codomain is surjective.  Just picking one function that's obviously not surjective does not prove that no other function is surjective.
Also, you should not say "Let $A$ be the set $\{a_1,a_2,a_3,\ldots\}$" unless you're talking about a countable set.  Cantor's theorem does not deal only with countable sets.  Moreover, even when the set is countable, referring to a specific way of counting it is just clutter if it's not actually used in the ensuing argument.
