Try a pointwise evaluation, i.e. $\delta_0:f\mapsto f(0)$.
Edit: A few clarifications.
First, $\delta_0$ is well-defined; if you want to see $C[0,1]$ inside $L^p$ already, then every time an equivalence class contains a continuous function, such function is unique.
As for why the extension cannot exist: first, one needs to say clearly what it means "doesn't exist". For linear functions can always be extended between vectors spaces (this is just linear algebra). The point here is that one wants a continuous extension. But the norm in $L^p$ is a different one, and sequences that didn't had a limit now do. For instance, consider the functions $g_n(t)=(1-nt)1_{[0,1/n]}$. They are all continuous and all satisfy $g_n(0)=1$, so $\delta_0(g_n)=1$ for all $n$. But, in $L^p$, $g_n\to0$, and thus $\delta_0(\lim g_n)\ne\lim\delta_0(g_n)$. So $\delta_0$ cannot be continuous.