Let $L^\infty = L^\infty(m)$, where $m$ is Lebesgue measure on $I=[0,1]$. Show that there is a bounded linear functional $\Lambda\neq 0 $ on $L^\infty$ that is $0$ on $C(I)$, and that therefore there is no $g\in L^1(m)$ that satisfies $\Lambda f = \int_I fg \, dm$ for every $f\in L^{\infty}$. Thus $(L^\infty)^*\neq L_1.$
We may see $C(I)$ as a subspace of $L^\infty$. Taking any $f\in L^\infty \setminus C(I),$ we consider the direct sum $\mathbb Rf \oplus C(I).$ There we may define the following linear functional $ \Phi(cf+g) = c$, which is null over the subspace $C(I)$. Now I want to apply the Hahn-Banach theorem to obtain a linear functional defined all over $L^\infty$. But for that, we must first ensure that $\Phi\leq p,$ where $p: L^\infty \rightarrow \mathbb R$ is a sublinear map. The most natural $p$ to take is $p = \|.\|_\infty$, but I am not being able to prove that $\Phi\leq p$.
Any help is much appreciated
Your approach will not work. The reason is that you need to let $p=||\cdot||_\infty$ in order for $\Phi$ to extend to a bounded linear functional. But this is impossible because $||cf+g||_\infty$can be bigger than $|c|$. For a counter-example, think of $f=\chi_{[0,1/2]}$ and choose $g\leq 0$ well.
To construct a functional that works, you should think more carefully about what is special about continuous functions. Your above idea doesn't use anything about them besides that they are a subspace. As a hint: