The family of all balls is a base for the topology. Fix a point $x$. By the definition of local compactness, there exists $V$ open with $x\in V$ and $\overline V$ compact. Then there is a ball $B_\delta(x)\subset V$. The closure of this ball is in $\overline V$, so it is compact. But then $B_1(0)=\frac1\delta\,B_\delta(x)-x$ is compact, too.
(a) The attempt works. To see that $S:=\{e_i,i\in\Bbb N\}$ is closed for the $\ell^1$ norm, let $x\notin S$. There are two index $i$ and $j$ such that $x_ix_j\neq 0$. Let $r:=\min\{|x_i|,|x_j|\}$. Then the open ball of center $x$ and radius $r$ is contained in the complement of $S$.
(b) The problem is that we have to check that we have convergence in $\ell^1$ for the subsequence.
As $\ell^1$ is complete, we can check that $B$ is precompact, i.e. given $\delta>0$, we can cover $B$ by finitely many balls of radius $<\delta$. It's equivalent to show both properties hold:
- $B$ is bounded in the $\ell^1$ norm;
- $\lim_{N\to +\infty}\sup_{x\in B}\sum_{k=N}^{+\infty}|x_k|=0$.
Indeed, if a set $S$ is precompact, with $\delta=1$ we get that it's bounded, and $2.$ is a $2\delta$ argument (I almost behaves as a finite set).
Conversely, assume that $1.$ and $2.$ hold and fix $\delta$. Use this $\delta$ in the definition of the limit to get an integer $N$ such that $\sup_{x\in B}\sum_{n=N+1}^{+\infty}|x_n|<\delta$. Then use precompactness of $[-M,M]^N$, where $M=\sup_{x\in B}\lVert x\rVert_1$.
Note that this criterion works for $\ell^p$, $1\leqslant p<\infty$.
In our case, each element of $B$ has a norm $\leqslant 1$, and for all $x\in V$,
$$\sum_{k=N}^{+\infty}|x_k|\leqslant \frac 1N\sum_{k= N}^{+\infty}k|x_k|\leqslant \frac 1N.$$
Best Answer
Let $e_n$ be $0$ except for a $1$ in the $n$th coordinate. Clearly $e_n \in c_0$. Then $\|e_n\|_\infty = 1$, but $\|e_n -e_m \|_\infty = 1$ whenever $n\neq m$. Hence $e_n$ can have no convergent subsequence, hence the set $\{ x \in c_0 | \|x\|_\infty \leq 1 \}$ cannot be compact.
And yes, all subsequences of $c_0$ must converge, since any subsequence of a convergent sequence must converge to the same limit.