Is the following argument correct?
Theorem. Every ordered field has countably infinite subset.
Proof. Let $F$ be an arbitrary ordered field. From definition $\textbf{1.1.5}$, we know that there exists a $1\in F$ such that $1\ne 0$, from the very same definition we also know that $x+y\in F$, whenever $x,y\in F$, implying $$A = \{1,1+1,1+1+1,\dots\}\subseteq F$$ To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $\textbf{1.1.8}$ shows that $x^2>0$ whenever $x\neq 0$, thus in particular $1>0$, definition $\textbf{1.1.7}$ clause $(i)$ together with definition $\textbf{1.1.1}$ clause $\textit{(ii)}$ implies that $$\cdots>1+1+1+1>1+1+1>1+1>1$$ appealing then to definition $\textbf{1.1.1}$ clause $\textit{(i)}$ then implies that $\sum_{i=1}^{j}1\neq\sum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.
$\blacksquare$
Note:
Definition $\textbf{1.1.5}$ is just the standard definition of a field.
Definition $\textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that
- $(i)$ For any $x,y,z\in F$ $x<y$ implies $x+z<y+z$
- $(ii)$ For any $x,y\in F$, $x>0$ and $y>0$ implies $xy>0$
Definition $\textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that
- $(i)$ For any $x,y\in S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.
- $(ii)$ If $x<y$ and $y<z$ then $x<z$.
The argument is pretty solid, nice work!
The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.