You use the Hahn-Banach theorem. Explicitly describing such a functional is very difficult but we can define a linear functional on a smaller space and extend it to the whole of $\ell^{\infty}$.
Let $X \subset \ell^{\infty}$ be the space of convergent sequences (it is easy to check this is a subspace). Consider the function $T': X \to \mathbb{R}$ given by $T'(x) = \lim_{n \to \infty} x_n$. Show that $T'$ is a bounded linear operator.
Then by Hahn-Banach $T'$ extends to a bounded linear operator $T: \ell^{\infty} \to \mathbb{R}$ (so $T \in (\ell^{\infty})^*$). Since $c_0 \subset X$, $T(x) = T'(x) = \lim_{n \to \infty} x_n = 0$ for all $x \in c_0$ therefore $T$ vanishes on $c_0$. However $T$ is not the zero operator since $T(1, 1, \ldots) = 1$.
You use the Hahn-Banach theorem. Explicitly describing such a functional is very difficult but we can define a linear functional on a smaller space and extend it to the whole of $\ell^{\infty}$.
Let $X \subset \ell^{\infty}$ be the space of convergent sequences (it is easy to check this is a subspace). Consider the function $T': X \to \mathbb{R}$ given by $T'(x) = \lim_{n \to \infty} x_n$. Show that $T'$ is a bounded linear operator.
Then by Hahn-Banach $T'$ extends to a bounded linear operator $T: \ell^{\infty} \to \mathbb{R}$ (so $T \in (\ell^{\infty})^*$). Since $c_0 \subset X$, $T(x) = T'(x) = \lim_{n \to \infty} x_n = 0$ for all $x \in c_0$ therefore $T$ vanishes on $c_0$. However $T$ is not the zero operator since $T(1, 1, \ldots) = 1$.