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."
Best Answer
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.