$\overline{B}_{1}(\textbf{0})$ complete $\iff (X,\|\cdot\|)$ Banach Prior to this, I have shown that $\overline{B_{1}}(\textbf{0}) = \overline{B_{1}(\textbf{0})}.$ Now,

Let $(X,\|\cdot\|)$ be a normed linear space. Show $\overline{B_{1}}(\textbf{0})$ is complete $\iff (X, \|\cdot\|)$ is a Banach space.

The reverse case I have as very simply: 

Suppose $(X,\|\cdot\|)$ is a Banach space. Thus, $X$ is complete. Since $\overline{B_{1}}(\textbf{0})$ is closed in $X$, it is complete. 

However, I am unsure of the reverse direction. I know that I want to show that $(X,\|\cdot\|)$ is complete. Take an arbitrary Cauchy sequence in $X$ and show that it converges in $X$ under $\|\cdot\|$. I am not sure how to continue.
 A: Let $\{x_n\}$ is Cauchy in $X$ and $\overline{B_1}(0)$ is complete. Because all Cauchy sequences are bounded it follows $\exists M>0:\,\sup\limits_{n\in\mathbb N}{\|x_n\|}\leq M$ (take $M\ge 1$). Denote $y_n=\frac{x_n}{M}$ and note that $\|y_n\|\leq 1\Rightarrow y_n\in \overline{B_1}(0)$. Now we show that $\{y_n\}$ is also Cauchy: Let $\epsilon>0$ is arbitrary and find $N\in\mathbb N$ such that $\|x_n-x_m\|\leq \epsilon,\,\forall n,m\ge N$
$$\|y_n-y_m\|=\|\frac{x_n}{M}-\frac{x_m}{M}\|=\frac{1}{M}\|x_n-x_m\|\leq \epsilon,\,\forall n,m\ge N$$
So $\{y_n\}$ is a Cauchy sequence in $\overline{B_1}(0)$ and therefore $\exists y\in \overline{B_1}(0):\,y_n\to y$ which means $\frac{x_n}{M}\to y$ or equivalently $x_n\to My\in X$.
Boundedness of $\{x_n\}$:
Let $\epsilon=1$ (arbitrarily taken) and find $N\in\mathbb N:\,\|x_n-x_m\|\leq 1,\,\forall n,m\ge N$. This implies $\|x_n-x_N\|\leq 1,\,\forall n\ge N$ and from triangle inequality it follows $\|x_n\|\leq \|x_N\|+1,\,\forall n\ge N$. So $\|x_n\|\leq \max\{\|x_1\|,\|x_2\|,...,\|x_{N-1}\|,\|x_N\|+1\},\,\forall n\ge 1$.
