$e_n \to 0$ weakly in $l^\infty$

Alternatively and directly:

Assume by contradiction that $(e_n)_{n}$ does not converge weakly to zero. Then there exists a $\epsilon >0$ and a functional $f\in l_\infty^\ast$ s.t. $|f(e_n)|\geq \epsilon$ for infinitely many $n\in\mathbb{N}$. By passing to that subsequence, we have that $|f(e_{n_k})|\geq \epsilon$ for all $k\in\mathbb{N}$. Let $\lambda_k :=\text{sign } f(e_{n_k})$ and set $x_N:=\sum_{k=1}^{N} \lambda_k e_{n_k}\in l_\infty$, we have that $\|x_N\|_\infty=1$ and $|f(x_N)|= \sum_{k=1}^N |f(e_{n_k})|\geq N\epsilon$.

Since we can do that for every $N\in\mathbb N$, we get a contradiction to the fact that $f\in l_\infty^\ast$.