I have a question about:schauder basis for $\ell_\infty$
I know that $\ell_\infty$ is not separable, therefore has no Schauder basis.However I cannot understand why the set $\{e_1, e_2, e_3, \dotsc \}$ where $e_1=(1,0,0,\dotsc), e_2=(0,1,0,0,\dotsc), \dotsc$ can not work as a Schauder basis.
One answer is that let $x=(1,1,1,\ldots)\in \ell_\infty$. The proof of that "$\ell_{\infty}$ has no Schauder basis by contradiction?
Question:
(1) The standard basis $\{e_n\}_{n\ge 1}$ is not Schauder basis
(2) What can we get from the results "$\sum_{I=1}^{\infty}x_ie_i$ does not converges to $x$"? I just read the proof in the second answer in schauder basis for $\ell_\infty$. I am confused about the answer.
Let $x=(1,1,1,\ldots)\in \ell_\infty$ if $x=\sum\limits_{i=1}^{\infty}x_ie_i,$ then for any $N\in\mathbb{N}$ we have $$\left\|x-\sum_{n=1}^{N}x_ne_n\right\|_{\infty}=\|(0,\ldots,0,1,1,1,\ldots)\|_{\infty}=1\not\to0,$$ so $\sum\limits_{i=1}^{\infty}x_ie_i$ does not converge to $x.$
The proof of (1): Does it make sense?
Suppse that $\{e_n\}$ is a Schauder basis. Then for every $x\in \ell_{\infty}$ we have $x=\sum_{I=1}^{\infty} x_ie_i$. That means as $n \to\infty$,$$0\le \sup_{k\ge n+1}|x_k|=\|(0,0,\dots,0, x_{n+1}, x_{n+2},\dots)\|_{\infty}=\|x-\sum_{I=1}^n x_ie_i\|_{\infty}\to 0$$
Thus, $x_k=0$ for $k\ge n+1$. So$$x=\sum_{I=1}^n x_ie_i$$
But this is impossible. Does it a contradiction?