Is this valid: Every Cauchy sequence in a normed space is absolutely convergent. Proof.  Let $X$ be a normed space with norm $|\cdot |$ and $(x_n)$ be Cauchy.  Then for all $\epsilon \gt 0, \ \exists N : m,n \gt N \implies |x_m - x_n| \lt \epsilon$ is the standard definition of Cauchy sequence.  But it's easy to show that $||x_n| - |x_m|| \leq |x_m - x_n|$ and thus the sequence $|x_n|$ in $\mathbb{R}$ is Cauchy.  By completeness of the reals under the absolute value norm, we have that $|x_n|$ approaches a limit and thus $(x_n)$ is absolutely convergent.
 A: The proof is correct. Applied more generally, it shows the following: 

If $X$ and $Y$ are metric spaces, $(x_n)$ is Cauchy in $X$, and $f: X\to Y$ is a uniformly continuous map, then the sequence $f(x_n)$ has a limit. 

The particular statement uses $Y=\mathbb R$ and $f(x)=\|x\|$ (which is a Lipschitz function). 

That said, I don't understand the bigger picture. Apparently "absolutely convergent sequence" here means a sequence $(x_n)$ such that $\|x_n\|$ has a limit. This is the first time I see this term used anywhere (and I kind of hope it's the last one. It seems designed to confuse people.) More importantly, this notion of "absolutely convergent sequence" does not imply usual convergence,  e.g., consider $x_n=(-1)^n$  in $\mathbb R$. And since $\mathbb R$ is a Banach space, this disproves the claim made in a comment, "a normed space is a Banach space iff absolutely convergent sequences converge". 
Perhaps the intended claim was "a normed space is a Banach space iff absolutely convergent series converge". That is indeed correct, but then the argument given in the OP is not really relevant. 
