The set of all real functions of a real variable 
How can I prove that the set of all real functions of a real variable,
  or even that the set of functions that take only the values 0 and 1,
  more than the continuum?


I have one idea, but it's not remarkable for its rigor and formality. I suppose that the power of all real functions is $\mathfrak{c}^\mathfrak{c}$ and the power of all functions that take only the values 0 and 1 -- $2^\mathfrak{c}$. So how $$\mathfrak{c}^\mathfrak{c} = 2^{\aleph_{0}\mathfrak{c}}=2^\mathfrak{c}>\mathfrak{c}.$$
Could you give more rigorous proof?
 A: Let $S=\{f:\Bbb R\to\{0,1\}\}$ and let $\alpha:\Bbb R\to S$ be surjective. Define for each $x\in\Bbb R$
$$f(x)=1-\alpha(x)(x)$$
Note that $f\in S$. Since $\alpha$ is surjective, there exists some $y\in\Bbb R$ such that $\alpha(y)=f$. Then
$$f(y)=1-\alpha(y)(y)=1-f(y)$$
which is a contradiction.
Remark: Note that $\Bbb R$, or even its cardinality, doesn't play any special role in the proof, so it could have been any nonempty set. Thus,

If $X$ is a nonempty set, the cardinal of $2^X$ is strictly greater than the cardinal of $X$.

The statement is also true for $X=\emptyset$, since $2^0>0$.
A: Let $S$ be some set.  Then $|2^S| > |S|$, where $2^S$ denotes the power set of $S$, and the set of functions from $S$ to ${0,1}$ has an obvious bijection with the power set of $S$.
Basically it is a one-step reduction to cantor's theorem -- the set of functions from $\mathbb{R}$ to $\{0,1\}$ is equal in size to the power set of $\mathbb{R}$.
$|2^S| > |S|$ is Cantor's theorem.  For empty $S$ check it manually.  For non-empty $S$:
Clearly $|2^S|\geq |S|$footnote 1.  So assume there is an surjection $f:S\to2^S$.
Define $Q_f := \{ x \in S : x \notin f(x)\}$, the Cantor diagonal set of $f$.
By surjectivity, $\exists y \in A : f(y) = Q_f$ (here is where I use non-empty).
Either $y \in Q_f$ or $y \notin Q_f$.
If $y \in Q_f$, then by construction of $Q_f$, $y \notin Q_f$.
If $y \notin Q_f$, then $y \notin f(y)$, and by definition of $Q_f$ we have $y \in Q_f$.
Only one assumption was made, and we have a contradiction, so there is no surjection from $S \to 2^S$.  Which means $2^S > S$.

1 $\{\{x\}\}$ is in $2^S$ for each $x \in S$
