How does one prove $a\gt b$ implies $ac \gt bc$ for $c\gt0$ when constructing the reals from equivalence classes of rational Cauchy sequences? I finally presented my problem more effectively in a comment below and have therefore erased the unsuccessful presentation of it that was formerly here.
 A: Claim : If  $a_n $ is a Cauchy sequences of rationals which is not equivalent to the zero sequences. Then  there  is a Cauchy sequence $b_n$ which is equivalent to $a_n$ and is bounded away from zero, i.e., there is some $c\in \mathbb{Q}^{>0}$ such that $|b_n|\ge c$ for all $n$.
Proof: Since $a_n $ is not equivalent to the zero sequence, there is some $\varepsilon>0$ such that for all $n_0\ge 0$ there is $n\ge n_0$ so that $|a_n|\ge \varepsilon$.  Also $(a_n)$ is Cauchy, so  $|a_i-a_k|< \varepsilon/2$ for all $i,k\ge N$. Choose $i\ge N$ such that  $|a_i|\ge \varepsilon$. Then 
$$|a_k|\ge |a_i|-|a_k-a_i|>\varepsilon/2$$
Hence for all $k\ge N$ we must have that $|a_k|>\varepsilon/2$. Define $b_k=a_k$ for all $k\ge N$ and $b_k = \varepsilon/2$ for $k<N$. Thus $b_n$ is equivalent to $a_n$ and  is bounded away from zero.  $\square$  
Now we define order in the reals as follows: We say that the real number $x$ is positive iff there exists a Cauchy sequence  of rationals $(x_n)$ such that $x_n \to x$, i.e, is in the class $[x]$ and $x_n$ is positively bounded away from zero for all $n$, i.e., there is a positive rational $c$ such that $x_n\ge c$ for all $n$ [of course we need to show that this is well-defined].   Because we know that $x$ is non-zero so is not equivalent to the zero sequence and for what we have shown there is a sequence which is bounded away from zero and once you get out as far as $N$, all terms must be on the same side of $0$. So you can define a equivalent sequence as above and you get a sequence of just positive terms.
Now we claim that if $x$ and $y$ are positive as reals, then $xy>0$. Since $x>0$ so there is a Cauchy sequence of rationals which is bounded away from zero and $x_n\to x$ similarly for $y$. Let $(x_n), (y_n)$ be such a sequences. So all the terms are positive as rational, we already know that $x_n y_n\ge c_1 c_2$ (notice that here we're talking about rational numbers), where $c_1,c_2$ are the positive lower bounds of $x_n,y_n$. Thus $\lim x_n y_n =xy\ge c_1 c_2>0$. Hence $xy>0$.
In your case $a>b$, which means $a-b>0$ and $c>0$ so $(a-b) \cdot c >0$, i.e., is positive real. If $a_n \to a, b_n \to b, c_n \to c$ and all are Cauchy sequences of rationals. Then $(a_n -b_n) c_n = a_n c_n-b_n c_n$ as rational and in the limit $ac-cb$  is positive and because the product and sum component-wise are well-defined (something that need to be proved) then $(a-b) \cdot c=ac-cb>0$,  so $ac>cb$.
