Complete subset of $X$ such that elements are of a fixed norm value implies $X$ Banach I am doing the exercise 5 in section 5.2 of Naylor's Linear Operator Theory in Engineering and Science.

Let $(X, ||·||)$ be a normed linear space and let $S_r = \{x ∈ X : ||x|| = r\}$
  where $r > 0$. Assume that $X \neq \{0\}.$ Show that $(X, || · ||)$ is a Banach space if and only if $(S_r, || · ||)$ is
  complete for some $r > 0$.

The left to right proof is easy enough. However, I'm having trouble proving right to left. In fact, I think it's incorrect. As a counterexample, let $X$ be the rationals over the field of rationals with the usual absolute value norm. This should be indeed a normed linear space. 
Fix $r$>0. So $S_r$ should just be $\{-r,r\}$. Here, the Cauchy sequences are the ones that are eventually constant. These clearly converge to either $-r$ or $r$. Hence $S_r$ is complete, but $X$, the rationals, is not Banach.
Am I missing something? Any help is appreciated.
Edit: As pointed out, my counterexample fails since my space must be over the reals or the complex numbers. This still leaves me at a roadblock. Any hint is appreciated.
 A: Hint
Suppose that $(S_r, \Vert \cdot \Vert)$ is complete and consider a Cauchy sequence $(x_n)$ in $X$. $(x_n)$ is bounded.
If $(x_n)$ converges to $0$, we are done. So we can suppose that $(x_n)$ doesn’t converge to $0$. This implies that it exists $d >0$ and a subsequence $(x_{\varphi(n)})$ such that $\Vert x_{\varphi(n)}\Vert \ge d$ for all $n \in \mathbb N$.
As $(x_n)$ is Cauchy, it exists $L \in \mathbb N$ such that for $s,t \ge L$ $\Vert x_s-x_t \Vert \le d/4$. Fix $m$ such that $M=\varphi(m) \ge L$. For $s \ge M$ $x_s$ belongs to the closed ball $B$ centered on $x_M$ of radius $d/4$.
Now $(\frac{r}{\Vert x_s \Vert})_{s \ge M}$ is a real sequence which is Cauchy. Hence converges to $l >0$. For $s,t \ge M$, you have
$$\begin{aligned}\left\Vert \frac{r}{\Vert x_s \Vert}x_s -\frac{r}{\Vert x_t \Vert}x_t  \right\Vert &= \left\Vert \left(\frac{r}{\Vert x_s \Vert}x_s - l x_s\right) + (lx_s - l x_t)  -\left(\frac{r}{\Vert x_s \Vert}x_s - l x_s\right) \right\Vert\\
& \le \left(\left\Vert x_s\right\Vert + \left\Vert x_s\right\Vert\right)\left\vert 
 \frac{r}{\Vert x_s \Vert} - l  \right\vert + l \Vert x_s - x_t \Vert
\end{aligned}$$
Inequality that you can use to prove that the sequence $(r \frac{x_s}{\Vert x_s \Vert})_{s\ge M}$ is a Cauchy sequence of $S_r$. That converges under the hypothesis that $S_r$ is complete.
Conclude by proving that this implies that $(x_s)_{s \ge M}$ converges.
A: Quote form the Wikipedia: "... a normed vector space is a vector space over the real or complex numbers..."
Hint for the solution to the problem:
$X\backslash\{0\} = \{tx:t > 0,x\in S_r\}$.
