Consider $(C[0,1];||.||_{\infty})$ and define the fuctional $f:\langle t\rangle \to \mathbb{R}$ by $$f(\alpha t)=\alpha.$$ It's easy to see that $||f||_{\infty}=1$, so by Hahn-Banach theorem I have a extension $F$ of $f$ to all $C[0,1]$ such that $||F||=1$.
But is this extension unique? (the only result involving uniqueness of Hahn-Banach Extensions I know is for strictly convex spaces and $C[0,1]^*$ isn't strictly convex).
Let $\varphi \in C([0,1])$ be any continuous function which vanishes on some interval $[1-\delta, 1]$. If $F(\varphi) \ne 0$, then by considering the function $\psi(t) = t \pm \epsilon \varphi(t)$ for sufficiently small $\epsilon$, we will have $\|\psi\|_\infty \le 1$ but $F(\psi) > 1$, contradicting $\|F\|=1$. So we conclude $F(\varphi) = 0$ for all such $\varphi$.
Now suppose that $\varphi \in C([0,1])$ with $\varphi(1) = 0$. We can find a sequence of continuous functions $\varphi_n$ vanishing on $[1-\delta_n, 1]$, with $\varphi_n \to \varphi$ uniformly. Then by the continuity of $F$ we conclude $F(\varphi)=0$ in this case also.
Finally let $u \in C([0,1])$ be arbitrary. Let $\varphi(t) = u(t) - u(1) t$, so that $u(t) = \varphi(t) + u(1) t$. Then $\varphi(1) = 0$ and so $F(\varphi) = 0$ as previously argued. By linearity of $F$ we conclude $F(u) = u(1)$.
So the only possibility for $F$ is $F(u)=u(1)$, and $F$ is indeed unique.