How can I show that given a norm one linear functional on $c_0$ that there is a unique extension to a norm one functional on $\ell_\infty$?

We are given that our Banach space is $c_0 \subset \ell_\infty(\mathbb{N})$ and there is a functional $y^* \in c_0^*$ such that $||y^*|| = 1$. We are guaranteed that this extends, via Hahn-Banach to a functional $x^* \in \ell_\infty^*$. How can I prove this extension is always unique in this case?

It is not true that such extension is unique. The assertion of uniqueness is the same as saying that if $x^*\in\ell^\infty(\mathbb N)^*$ and $x^*|_{c_0}=0$, then $x^*=0$. That's not the case.

We can construct counterexamples by noting that $\ell^\infty(\mathbb N)=C(\beta\mathbb N)$, where $\beta \mathbb N$ is the Stone-Cech compactification of $\mathbb N$. The elements in $\beta\mathbb N\setminus\mathbb N$ are usually identified with the free ultrafilters. For any $\omega\in\beta\mathbb N\setminus\mathbb N$, we may consider the functional $\varphi_\omega:\ell^\infty(\mathbb N)\to\mathbb C$ given by $$\varphi_\omega(x)=x(\omega).$$ Being a point-evaluation, $\|\varphi_\omega\|=1$.

The number $x(\omega)$ is often denoted by $\lim_{n\to\omega} x(n)$. It is actually a limit, in the sense that $$\tag1x(\omega)=L\iff\ \text{ for every }\varepsilon>0,\ \{n:\ |x(n)-L|<\varepsilon\}\in\omega.$$ As a free ultrafilter contains all the sets $X_n=\{k:\ k\geq n\}$, it is easy to see that if $\lim_{n\to\infty}x(n)=L$, then $x(\omega)=L$ too. In particular, for any free ultrafilter $\omega$ we have $x(\omega)=0$ for all $x\in c_0$.

So to contradict the uniqueness it is enough to find $x\in\ell^\infty(\mathbb N)$ and $\omega_1,\omega_2\in\beta\mathbb N\setminus\mathbb N$ such that $x(\omega_1)\ne x(\omega_2)$. We may take $x(n)=(-1)^n$, and $\omega_1$ and $\omega_2$ respectively the free ultrafilters generated by $2\mathbb N$ and $1+2\mathbb N$ respectively. These two ultrafilters are distinct since an ultrafilter that contains a set does not contain its complement. And its easy to see from $(1)$ that $$x(\omega_1)=1\ne-1=x(\omega_2).$$