Prove X is banach iff $S(0,1)$ is complete. Let $X$ a normed space. Let $S(0,1)=\{x\in X: ||x||=1\}$ Prove that $X$ is banach iff $S(0,1)$ is complete.
My attempt:
$(\implies)$ Note $S(0,1)\subset X$ and we have $X$ is banach space. 
Let $\{x_n\}$ a cauchy sequence in $S(0,1)$. We need prove that $\{x_n\}$ converge.
As $\{x_n\}$ is cauchy then exists $N$ such that if $n,m>N$ then $||x_n-x_m||<\epsilon$
Here i'm stuck.
$(<-)$  Let $\{x_n\}$ a cauchy sequence in $X$.
Note for $n,m>N$ we have $||x_n-x_m||<\epsilon$.
 A: One direction is clear. If $X$ is Banach, then $S(0,1)$ being a closed subset (inverse image of $\{1\}$ under the continuous map $x\mapsto\|x\|$ from $X\to\mathbb{R}$), is also complete.
For the converse part, assume that $X$ is a normed linear space such that $S(0,1)$ is complete. Let $(x_n)$ be a Cauchy sequence in $X$. This tells us that $\|x_n\|$ is a Cauchy sequence in $\mathbb{R}$. 
Case 1: Suppose $\|x_n\|\to 0$.
In this case by definition of convergence, we can see that $x_n\to 0$.
Case 2: $\|x_n\|\not\to 0$.
In this case we can find two positive constants $\alpha,\beta$ such that $\alpha\leq\|x_n\|\leq\beta$ for all $n\in\mathbb{N}$. (We are using the fact that Cauchy sequences in $\mathbb{R}$ are bounded).
Without loss of generality we can also assume that $x_n\neq 0$ for all $n\in\mathbb{N}$ because there will be at most finitely many of such elements. This is because, if there were infinitely many of them, we would have a subsequence of $(x_n)$ which is converging to $0$. Thus, $(x_n)$ being Cauchy, we will get that $(x_n)\to 0$ which we have assumed otherwise.
Now define $y_n=\frac{x_n}{\|x_n\|}$. (This is well defined because we have assumed $x_n\neq 0$ for all $n\in\mathbb{N}$). Therefore we have,
$$\|y_n-y_m\| = \left\|\frac{x_n}{\|x_n\|} - \frac{x_m}{\|x_m\|}\right\|\leq \left\|\frac{\|x_m\|x_n-\|x_n\|x_m}{\|x_n\|\|x_m\|}\right\|\leq \frac{1}{\alpha^2} \left\|\|x_m\|x_n-\|x_m\|x_m+\|x_m\|x_m-\|x_n\|x_m\right\|\leq\frac{1}{\alpha^2} \left(\|x_m\|\|x_n-x_m\| + \|x_m\|\cdot\left|\|x_m\|-\|x_n\|\right|\right)\leq \frac{\beta}{\alpha^2} \left(\|x_n-x_m\| +\left|\|x_m\|-\|x_n\|\right|\right).$$
Now since we have assumed $(x_n)$ to be Cauchy (in $X$) and we have seen that $\|x_n\|$ is Cauchy (in $\mathbb{R}$), we can see that $(y_n)$ is Cauchy. Therefore it converges in $S(0,1)$ by assumption. Let's say it converges to $y$. If we assume that $\|x_n\|\to a\in\mathbb{R}$, then we can see that $x_n=\|x_n\| y_n$ converges to $a\cdot y$. Thus proving that $X$ is a Banach space.
A: $(\Rightarrow)$ Let $\{x_n\} \subset S(0,1)$ a Cauchy sequence. Since $S(0,1) \subset X$ and $X$ is Banach, exists $x \in X$ such that $x_n \to x$. 
Affirmation: $x \in S(0,1)$
In fact, given $\epsilon >0$, exists $n_0 \in \mathbb{N}$ such that:
$$
n \ge n_0 \implies ||x_n-x|| < \epsilon
$$
By triangular inequality:
$$
||x|| \le ||x- x_n|| + ||x_n|| < 1 + \epsilon \ \mbox{and} \ 1 = ||x_n||\le ||x_n-x|| + ||x|| \implies 1 - \epsilon < ||x||
$$
That is:
$$
1 - \epsilon < ||x|| < 1+ \epsilon
$$
Doing $\epsilon \to 0^{+}$, we get that $||x|| = 1 \implies x \in S(0,1)$. Then, $S(0,1)$ is complet.
$(\Leftarrow)$ Let $\{x_n\}\subset X$ a Cauchy sequence. Define for $x_n \neq 0$, $y_n = \dfrac{x_n}{||x_n||}$. We have to $\{y_n\}$ is a Cauchy sequence in $S(0,1)$. Since $S(0,1)$ is complete, exists $y\in S(0,1)$ such that $y_n \to y$. Observe that $\{||x_n||\}\subset \mathbb{R}$ is a Cauchy sequence. Thus, exists $a\in \mathbb{R}$ such that $||x_n||\to a$. Define $x=ay$.
Affirmation: $x_n \to x$.
In fact, given $\epsilon > 0$
$$
||x_n - x|| = ||(||x_n||)y_n-ay|| = || (||x_n||)y_n - ay_n + ay_n - ay || \le ||y_n||[|||x_n||-a|] + a||y_n-y|| \le [|||x_n||-a|] + a||y_n-y|| < \epsilon
$$
for $n$ big enough. Therefore $x_n \to x \implies X$ is Banach. 
