The dual to the space of continuous functions is not isomorphic to $\ell^\infty$ This is a follow-up to this answer, which in turn arose in a follow-up to this question.
Let $M([0, 1])$ denote the Banach space of signed Borel measures on $[0,1]$ equipped with the total variation norm. This space is the dual of $C([0, 1])$, via the obvious identification. 

Question. Is $M([0, 1])$ isomorphic to $\ell^\infty$?

I expect the answer to be negative.
 A: There are several ways to proceed depending on how much machinery you are willing to use.
$\ell_\infty$ is a von Neumann algebra and thus, by Sakai theorem, it has a unique predual. Nevertheless $M([0,1])$ has plenty of nonisomorphic preduals, for instance $M([0,1])$ and $M([0,1]^2)$ are isomorphic while $C([0,1]^2)$ and $C([0,1])$ are not.
In the particular case of $\ell_\infty$ you can see that it has a unique predual directly. Use that $E^\ast = X_1 \oplus_\infty X_2$ implies that $E = E_1 \oplus_1 E_2$ with $E_1$ and $E_2$ the preduals of $X_1$ and $X_2$ respectively. Iterating will give you that $E$ satisfying that $E^\ast = \ell_\infty$ has a nested family $\ell^1(1) \subset \ell_1(2) \subset ... \ell_1(N)$ dense in $E$. Therefore $E \cong \ell_1$.
There other strategies, what I sugested in the comments was using injectivity.
A Banach space $E$ is called injective iff it satisfies a Hanh-Banach theorem if you use as an endpoint space. I.e.: for every $Y \subset X$ and bounded map $\phi: Y \to E$ there is a bounded extension $\tilde{\phi}: X \to E$ with the same norm.
Observation $\ell_\infty$ is trivially injective because we can apply Hanh-Banach in each of the coordinates. 
I think, although I do not have a reference at hand, that all injective Banach spaces are $1$-complemented subsets of spaces of the form $C(K)$, where $K$ is totally disconnected. 
It is easy to check that a Banach space is injective iff whenever it sits inside a larger space, it is complemented. Therefore it is enough to see that $M([0,1])$ sits as a subset of a larger space in a way that is not complemented. Probably there are examples of this in the literature, but I cannnot came up with one. The closest thing I can think of is:


*

*Use that $M([0,1]) \cong \mathbb{C} \oplus_1 \mathbb{C} \oplus_1 M(0,1)$, to reduce the problem to the open interval.

*Since $(0,1)$ is homeomorphic to $\mathbb{R}$, $M(0,1)$ is isomorphic to $M(\mathbb{R})$.

*Take $M(\mathbb{R}) \subset L^1(\mathbb{R})^{\ast \ast}$, the double dual of $L^1$. Assume that there is a projection $P: L^1(\mathbb{R})^{\ast \ast} \to M(\mathbb{R})$ to reach contradiction.

*The unit ball $B$ of $B(L^1(\mathbb{R})^{\ast \ast}, M(\mathbb{R}))$ is comapct in the point weak-$\ast$ topology given by the predual $M(\mathbb{R}) = C_b(\mathbb{R})/n$, where $n$ is the preannihilator of $M(\mathbb{R})$. The projections are a closed convex subset $B_0 \subset B$. There is an action of $\mathbb{R}$ over $B_0$ given by $t \mapsto \tau_t \circ P \circ \tau_{-t}$ where $\tau_t$ is the translation. By amenability, we have a fixed point $P_0 \in B_0$ which would be an equivariant projection (a projection commuting with the translation action). If the original $P$ would preserve $1_{\mathbb{R}}$, i.e.
$$\langle P(\psi),1_{\mathbb{R}} \rangle = \langle \psi, 1_{\mathbb{R}} \rangle$$ we would get a contradiction right away since an invariant mean $m \in L^1(\mathbb{R})^{\ast \ast}$ would give a $\mathbb{R}$-invariant and nonzero finite measure $P(m)$. I do not know whether you can take $P$ preserving $1$ without loss of generality.

