Banach spaces and their unit sphere

2) One direction is straightforward, the other one requires more work.

If $X$ is a Banach space, it is complete. Now $S$ is closed in $X$, so it is complete.

Conversely, assume that $S$ is complete. Take $(x_n)$ a Cauchy sequence in $X$. If $\lim x_n=0$, we are done proving that $(x_n)$ converges in $X$. So assume that $(x_n)$ does not converge to $0$. This means that there exists $\epsilon>0$ and a subsequence $(x_{n_k})$ such that $\|x_{n_k}\|\geq \epsilon$ for all $k$. Then write...

EDIT Thanks to @Jesper for pointing out a mistake...

$$ \frac{x_{n_k}}{\|x_{n_k}\|}-\frac{x_{n_l}}{\|x_{n_l}\|}=\frac{x_{n_k}-x_{n_l}}{\|x_{n_k}\|}+(\|x_{n_l}\|-\|x_{n_k}\|)\frac{x_{n_l}}{\|x_{n_l}\| \| x_{n_k}\|}. $$ It follows, applying triangular, reverse triangular inequalities and the fact that $\|x_{n_k}\|\geq \epsilon$ in the preceding equality that: $$ \left\| \frac{x_{n_k}}{\|x_{n_k}\|}-\frac{x_{n_l}}{\|x_{n_l}\|}\right\|\leq \frac{2}{\epsilon}\|x_{n_k}-x_{n_l} \|. $$ So the sequence $\frac{x_{n_k}}{\|x_{n_k}\|}$ is Cauchy in $S$. Therefore it converges to some $y$ in $S$.

Now note that $\|x_n\|$ is a Cauchy sequence again by the reverse triangular inequality as $|\|x_n\|-\|x_m\|| \le \|x_n - x_m\|$ and $(x_n)$ is Cauchy. Since $\mathbb{R}$ is complete, it follows that $\|x_{n_k}\|$ converges to $M$. Whence $x_{n_k}=\|x_{n_k}\|\cdot\frac{x_{n_k}}{\|x_{n_k}\|}$ converges to $My$. It only remains to apply 1) to see that $(x_n)$ converges. Hence $X$ is complete, i.e $X$ is a Banach space.