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I was told that the answer I gave to this question didn't actually show that the unit ball wasn't closed. Could someone please help me figure out how to do this? this is the original answer I gave ($S$ refers to the unit ball).

"We consider $S$ as stated. Clearly, $e_{i} \in \ell^{2}$ and $\|e_{i}\|_{2} = 1$ for all $e_{i}$, so $S$ is bounded and closed. .But since for $i \neq j$ $\|e_{i} - e_{j}\| = \sqrt{2}, $ the sequence $(e_{i})_{i=1}^{\infty}$ can not possibly have any convergent sub-sequences as they will not be Cauchy, so it is not compact."

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The inequality that $\big|\|x_{n}\|-\|x\|\big|\leq\|x_{n}-x\|$ shows that $\|x_{n}\|\rightarrow\|x\|$ if $\|x_{n}-x\|\rightarrow 0$. Now assume that $(x_{n})\subseteq S$ and $\|x_{n}-x\|\rightarrow 0$, then $\|x_{n}\|\rightarrow\|x\|$. But $\|x_{n}\|\leq 1$, so taking limit we have $\|x\|\leq 1$, so $x\in S$, $S$ is then closed.

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The set $S$ is closed because it is equal to $f^{-1}\bigl([0,1]\bigr)$, where $f(x)=\|x\|$. Since $f$ is continuous, this is enough.

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In any metric space $(X,d),$ for any $x\in X$ and any $r\in \Bbb R^+,$ the set $S=\{y\in X: d(y,x)\leq r\}$ is closed because its complement is open: If $y\in X\setminus S$ and $p=d(y,x),$ let $p'=(p+r)/2.$ The $\triangle$ inequality implies that the open set $B_d(y,p')=\{z\in X: d(z,y)<p'\}$ (which contains $y$) is disjoint from $S.$

A normed linear space is a metric space with the metric $d(x,y)=\|x-y\|.$

In a normed linear space we can also show that $\{x: \|x\|\leq 1\}$ is closed by using the basic result that if $\lim_{n\to \infty}\|x_n-x\|=0$ then $\|x\|=\lim_{n\to \infty}\|x_n\|.$ So if $\|x_n\|\leq 1$ for all $n,$ and if $(x_n)_n$ converges to $x,$ then $\|x\|=\lim_{n\to \infty}\|x_n\|\leq 1.$

You are correct when showing non-compactness.

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