Assuming that we have the definition of the rational numbers $\mathbb{Q}$ we define $\mathbb{R}$ as the completion of $\mathbb{Q}$ with respect to the usual norm. I want to understand why this gives a field.
By definition of the construction, $\mathbb{R}$ is defined as equivalence classes of Cauchy sequences $\{x_n\}$ with $x_n\in \mathbb{Q}$. That is $$ \mathbb{R} = \{[\{x_n\}] : \{x_n\}\text{ Cauchy in } \mathbb{Q}\}. $$ Here $\{x_n\} \sim \{y_n\}$ if $\lim_{n\to \infty} \lvert x_n - y_y\rvert = 0$.
I believe the operations on $\mathbb{R}$ are $$ [\{x_n\}] + [\{y_n\}] = [\{x_n + y_n\}] \\ [\{x_n\}] [\{y_n\}] = [\{x_ny_n\}] $$
I believe I have shown that $\mathbb{R}$ is a commutative ring. All I have left to show is that all the (non-zero) elements have multiplicative inverses.
But if I have $[\{x_n\}]\neq 0$, then I can just define $y_n = 1/ x_n$ so that $$ [\{x_n\}] [\{y_n\}] = [\{x_n y_n\}] = [\{1\}] $$
I am under the impression that the inverse part of the proof is hard, so I am guessing that there is something wrong with my approach.
Am I doing the right thing?
Let's see what's the meaning of $[\{x_n\}]\ne0$. Since $0$ is the equivalence class of the constant zero sequence, we know that it doesn't hold $$ \lim_{n\to\infty}|x_n|=0 $$ Thus there exists $\eta>0$ (rational) such that, for every $n$, there is $m>n$ with $|x_m|\ge2\eta$.
On the other hand, $\{x_n\}$ is Cauchy, so there exists $\bar{n}$ such that, for $m,n>\bar{n}$, $|x_n-x_m|<\eta$.
Fix $m>\bar{n}$ such that $|x_m|\ge\eta$; then, for every $n\ge\bar{n}$, $$ x_m-\eta<x_n<x_m+\eta $$ If $x_m>0$, then $x_m\ge2\eta$, so $x_n>\eta$; if $x_m<0$, then $x_m<-2\eta$ and so $x_n<-\eta$.
In either case we can state that $x_n\ne0$, for every $n>\bar{n}$.
Define a new sequence $\{y_n\}$ by $y_n=0$ if $0\le n\le\bar{n}$ and $y_n=x_n^{-1}$ if $n>\bar{n}$.
Now the task is to prove that $\{y_n\}$ is Cauchy. Once you have this, you can easily prove that $[\{x_n\}][\{y_n\}]=1$ (the equivalence class of the constant one sequence).
You need to use that $|x_n|\ge\eta$ for every $n>\bar{n}$.