Closed $\iff$ weakly closed subspace

Edit: the previous answer was wrong. Below is a new version.

Weak convergence of a net $\{x_n\}$ to $1\in\ell^\infty$ means that, for any functional $\varphi$ in the dual of $\ell^\infty$, $\varphi(x_n-1)\to0$.

We can see $\ell^\infty$ as $C(\beta\mathbb N)$, the continuous functions on the Stone-Čech compactification of $\mathbb N$. Let $\omega\in\beta\mathbb N\setminus\mathbb N$ (in other words, $\omega$ is a free ultrafilter). Then $1(\omega)=1$ and $x(\omega)=0$ for all $x\in c_0$. So we have, letting $\varphi_\omega$ be the point-evaluation at $\omega$, $$ \varphi_\omega(x-1)=\varphi_\omega(x)-\varphi_\omega(1)=0-1=-1, $$ and so no net in $c_0$ will make the limit go to zero. This means that $c_0$ is not weakly dense in $\ell^\infty$. Weak-star density works because one has to deal with less functionals.

As was mentioned, the Hahn-Banach theorem guarantees that $c_0$ (being convex) is both norm and weakly closed.

$c_0$ is not weakly dense in $\ell_\infty$. Indeed, let $u:=\sum_{k=0}^{+\infty}e^{2k}$, where $e^k_j:=\delta_{kj}$. Take a Banach limit $L$. Then $$U:=\{x\in\ell_\infty, |L(x-u)|<1/3\}$$ is a neighborhood of $u$ in the weak topology. If $x\in U\cap c_0$, then $L(x)=0$, hence $|L(u)|<1/3$. It's a contradiction, since $L(u)=1/2$.