How to prove $\mathbb{N}\subseteq\mathbb{R}$ given basic field axiomatisation. I understand that $\mathbb{R}$ can be defined axiomatically in a number of different ways, including as an extension from more basic number systems (e.g., $\mathbb{N}$, $\mathbb{Z}$, or $\mathbb{Q}$) using equivalence classes or Dedekind cuts. However, my question here assumes that $\mathbb{R}$ is defined as a complete ordered field directly from the following axioms:


Field Axioms (defining addition and multiplication with algebraic field properties -- in particular including a multiplicative constant $1\in\mathbb{R}$)
Order Axioms ($\mathbb{R}$ is totally ordered)
Completeness Axiom (the order is Dedekind-complete)


And now to the question: Given these axioms, how does one prove the existence of a Peano system $P\subseteq\mathbb{R}$ with $1_{\mathbb{R}}=1_P$ (i.e., I want to prove that $\mathbb{N}\subseteq\mathbb{R}$)?
With $S\colon \Bbb R\to\Bbb R$, $x\mapsto x+1_{\Bbb R}$ let $\Bbb N$ be the intersection of all subsets of $\Bbb R$ that contain $1_{\Bbb R}$ and are closed under $S$. 

That is: With $e=1_{\Bbb R}$ (or $e=0_{\Bbb R}$ depending on taste) the definition
$$\Phi(A)\stackrel {\text{def}}\iff  e\in A\land \forall x\in A\colon S(x)\in A$$
we let
$$\Bbb N\stackrel {\text{def}}=\bigcap\{\,A\in\mathcal P(\Bbb R)\mid \Phi(A)\,\}.$$
(This definition makes sense because such subsets $A$ do exist, for example we trivially have $\Phi(\Bbb R)$).
One verifies that $\Phi(\Bbb N)$, so that $(\Bbb N,e,S)$ is a viable candidate: $\Bbb N$ is a set, $e\in\Bbb N$ is an element, and $S\colon \Bbb N\to\Bbb N$ is a map. The axioms are readily verified:


Assume $e=S(a)$ for some $a\in\Bbb N$. From $\Phi(\{\,x\in\Bbb R\mid x\ge e\,\})$ (which follows from $1_{\Bbb R}\ge 0_{\Bbb R}$) we conclude that $a\ge e$, hence $e=a+1_{\Bbb R}\ge e+1_{\Bbb R}$, contradiction
Assume $S(a)=S(b)$. Then $a+1_{\Bbb R} = b+1_{\Bbb R}$, hence $a=b$.
Let $A\subset\Bbb N$ be a set with $e\in A$ and $\forall x\in A\colon S(x)\in A$. In other words, $\Phi(A)$. Then by definition $\Bbb N\subseteq A$, hence $A=\Bbb N$.


Note that we didn't use completeness (so we find $\Bbb N$ also inside $\Bbb Q$) or Archimedean-ness of the order. Ordered field is enough.
Hint: 

Along the lines of Hagen's answer, construct function $S$ such that: 

$$\forall a\in R: S(a)=a+1$$

Construct $N$ such that:

$$\forall a:[a\in N \iff a\in R \land \forall b\subset R:[0 \in b \land \forall c\in b: [S(c)\in b] \implies a\in b]]$$

Then prove:


$0\in N$
$\forall a\in N: S(a) \in N$
$\forall a,b \in N:[S(a)=S(b)\implies a=b]$
$\forall a\in N: S(a)\neq 0$
$\forall P\subset N: [0\in P \land \forall b\in P: [S(b) \in P] \implies P=N]$