$\mathbb R$ is uncountable I'm trying to prove that the real numbers are uncountable using two simple facts which I've already proved:
Fact 1

There aren't any surjection $f:\mathbb N\to P(\mathbb N)$

Fact 2

There exists an injective function $g:P(\mathbb N)\to \mathbb R$.

I would like to know if with only these facts I can deduce that the reals aren't countable.
Thanks in advance
EDIT
Following the comments, can I say the statement is true because of $\mathbb N\lt P(\mathbb N)\le \mathbb R$, where the first inequality comes from fact 1 and second inequality comes from fact 2.
 A: Suppose there exists a bijection $\varphi : \mathbb{N} \rightarrow \mathbb{R}$, and let $g : P(\mathbb{N}) \rightarrow \mathbb{R}$ be an injective map. Choose a left inverse of $g$, say $h : S \rightarrow P(\mathbb{N})$ where $S \subset \mathbb{R}$. Then we have a surjective map $h \circ \varphi : T \rightarrow P(\mathbb{N})$, where $T \subset \mathbb{N}$, $T = \varphi^{-1}(S)$, a contradiction.
A: Yes, your two facts are sufficient.
You can prove that for any two sets $A$ and $B$, there exists an injection $A\hookrightarrow B$ if and only if there is a surjection $B\twoheadrightarrow A$. You can also prove (the Cantor-Bernstein-Schroeder theorem) that if there are injective maps $A\hookrightarrow B$ and $B\hookrightarrow A$, then there is a bijection $A\leftrightarrow B$. From this, you can use the existance of an injection $A\hookrightarrow B$ to define $|A|\le|B|$, with $|A|=|B|$ if $|A|\le|B|$ and $|B|\le|A|$ (i.e. if there is a bijection between them), and defined $|A|<|B|$ as $|A|\le|B|$ but not $|A|=|B|$.
Finally, for any two sets $A$ and $B$, you can always find an injection $A\hookrightarrow B$ or $B\hookrightarrow A$.
Thus, your two facts mean that $|\mathbb{N}|<|P(\mathbb{N})|$ and $|P(\mathbb{N})|\le|\mathbb{R}|$.
A: Proposition: $\mathbb{R}$ is uncountable. 
Proof: Let define the map $f: \mathcal{P}(\mathbb{N}) \to \mathbb{R}$ by the formula $f(A):= \sum_{n\in A}10^{-n}$. Note that the map is well-define since $\sum_{n=0}^\infty 10^{-n}$  is absolutely convergent. 
We will show that $f$ is one-to-one. Suppose to the contrary that there exists $A, B \in \mathcal{P}(\mathbb{N}) $ such that $f(A)=f(B)$ but $A \not=B$. Then the set $(A\setminus B )\cup (B\setminus A)$ is non-empty. Let $n_0$ be the least element and for sake of definiteness assume that $n_0 \in A\setminus B $. So we have 
\begin{align}0=f(A)-f(B)=\sum_{n\in A}10^{-n}-\sum_{n\in B}10^{-n}=\sum_{n\in A:n<n_0}10^{-n}+10^{-n_0}+\sum_{n\in A:n>n_0}10^{-n}\\-\sum_{n\in B:n<n_0}10^{-n}-\sum_{n\in B:n>n_0}10^{-n}\\
=10^{-n_0}+\sum_{n\in A:n>n_0}-\sum_{n\in B:n>n_0}10^{-n}\\
\ge 10^{-n_0}-\sum_{n>n_0}10^{-n}\\
=10^{-n_0}-\frac{10^{-n_0}}{9}>0
\end{align}
a contradiction. Hence $f$ is an injection. 
Then $\mathcal{P}(\mathbb{N})\simeq f(\mathcal{P}(\mathbb{N}))$, i.e.,  $f(\mathcal{P}(\mathbb{N}))$  has the same cardinality as $\mathcal{P}(\mathbb{N})$ and by the Cantor's theorem  $\mathcal{P}(\mathbb{N})$ is uncountable. Since $f(\mathcal{P}(\mathbb{N}))$ is a subset of $\mathbb{R}$ then this forces $\mathbb{R}$ to be uncountable. 
