Consider the linear subspace $c$ of $l^\infty$ of those sequences $\{a_k\}_k$ which have a finite limit as $k \to +\infty$. Define a linear functional $T: c \to \mathbb{R}$ as follows: $$T(\{a_k\}_k)=\lim\limits_{k \to +\infty} a_k$$ This is an element of the dual space of $(c, ||\cdot||_{l^\infty})$.
I can see that its norm is $\le1$ because $$|T(\{a_k\}_k)|\le1\cdot||\{a_k\}_k||_{l^\infty}$$ How I can check that it is equal to $1$?
Moreover if I apply $T$ to the elements $e^h$ of the "canonical basis", obviously for each fixed $h$ we have $$T(e^h)=0$$ I think this becomes from the fact that a basis vector has all zeros in all positions except in the $h\mathrm{-th}$ position where we have a $1$, so the $k\mathrm{-th}$ term must be $0$ as $k \to +\infty$. Right?
The norm is $1$ because if $(\forall k\in\mathbb{N}):a_k=1$, then $\bigl\|\bigr(a_k)_{k\in\mathbb N}\|_\infty=1$ and $\lim_{k\to\infty}a_k=1$.
What you wrote about the $e^h$'s is correct, but don't call “canonical basis” to the set of all $e^h$'s, because they are not a basis of $c$.