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.
Consider the set of natural numbers $\mathbb N$. Let $d$ be the discrete metric, so that $d(m,n)=1$ if $n \neq m$ and $d(n,n)=0$. Define an alternative metric
$$\rho(m,n)=\left\vert \frac{1}{m} - \frac{1}{n} \right\vert.$$
Both $d$ and $\rho$ induce the discrete topology on $\mathbb N$, and hence are equivalent in the sense that they have the same convergent sequences. More explicitly, say that the metrics $d$ and $\rho$ are equivalent if $x_n \overset{d}{\to}x \iff x_n \overset{\rho}{\to} x$. This is the same as the metrics inducing the same topology.
However, the sequence $\left\{1,2,3,... \right\}$ is $\rho$-Cauchy but not $d$-Cauchy.
To see that $\rho$ induces the discrete topology, let $r_n = \frac{1}{n(n+1)}$, and observe that $B_{r_n}(n) = \{ n \}$.
Best Answer
Using a Hamel basis of an infinite dimensional Banach space $(X,\|\cdot\|_2)$ one finds a discontinuous linear functional $f$. We define a strictly stronger norm $\|x\|_1=\|x\|_2+|f(x)|$. The open mapping theorem implies that $(X,\|\cdot\|_1)$ is not complete but every $\|\cdot\|_1$-Cauchy-sequence is also $\|\cdot\|_2$-Cauchy and hence $\|\cdot\|_2$-convergent.