Prove that there is no injection from $\Bbb{R}$ to $\Bbb{N}$. Knowing that there is no surjection from $\Bbb{N}$ to $\Bbb{R}$, I also know that there is no injection going the other way.
I know how to prove that there is no injection from the power set of $S$ to $S$, but how can I prove that there is no injection from $\Bbb{R}$ to $\Bbb{N}$?
(I would, of course, be happy with showing no injection from $(0,1)$, or any other interval to $\Bbb{N}$.)
[EDIT]
I usually prove that cardinality of $\mathcal{P}(S)$ is greater than that of $S$ by showing that any function $f: S \to \mathcal{P}(S)$ fails to be surjective. This is done by showing that there is no $s \in S$ for which $f(s) = {\{ s\in S | s \notin f(s) \}}$.
It is possible instead to show that any function $f:\mathcal{P}(S) \to S$ fails to be injective by showing that the image of $T = \{s \in S| \exists U \left( f(U)=s \land s \notin U \right) \}$ must be the image of two elements of $\mathcal{P}(S)$.
I would like to replace the usual diagonal proof that any function $\Bbb{N} \to (0,1)$ fail to be surjective with a proof that any function $f:(0,1) \to \Bbb{N}$ fails to be injective.  (Or that any function $f:\Bbb{R} \to \Bbb{N}$ fails to be injective.)
 A: Schröder–Bernstein theorem:

If there exist injective functions $f : A \to B$ and $g : B\to A$ between the
  sets $A$ and $B$, then there exists a bijective function $h : A \to B$.

Your $A$ and $B$ are $\Bbb{N}$ and $\Bbb{R}$.
A: "I know how to prove that there is no injection from the power set of $S$ to $S$, but how can I prove that there is no injection from $\mathbb R$ to $\mathbb N$?"
Using binary decimal expansions of real numbers, you can show that there is an injection $(0,1)$ to $\mathcal P(\mathbb N)=2^\mathbb N$ with image whose complement is a countable set. Using the elementary fact you have above, this should be enough information to proceed directly, without appealing to the Cantor-Shroeder-Bernstein theorem.
UPDATE: I see your edit, and I think this answers your question better. Any injection $(0,1) \rightarrow \mathbb N$  will admit a partial inverse $M\rightarrow (0,1)$ where $M$ is some subset of $\mathbb N$. Then you can use that any infinite subset of $\mathbb N$ is in bijection with $\mathbb N$ (or prove from first principles of induction using the fact that $\mathbb N$ is well ordered). Then use your previous knowlegde of the diagonal proof that $\mathbb N$ does not inject into $(0,1)$, after eliminting the case that $M$ is finite.
Even if you want a direct proof, I believe some abstraction is probably helpful and the 'right' thing to do.
