The Banach space $C[0,1]$ is not reflexive. 
Equip the space $C([0,1])$ with the usual supremum norm. Show that this space is not reflexive.

I already saw a lot of solutions to this kind of question, but I just don't know how to start this.
Many people are starting with the fact "if $E$ is a reflexive Banach space then each linear continuous functional attains its norm." What does this mean? Or why is that true?
I just know that $X$ is reflexive if $J(X) = X''$ and $J$ is defined as $J\colon X \to X''$ by $J(x)=F_x$ and $F_x$ is the functional from $X \to F$.
 A: If $C([0,1])$ was reflexive then $B := \overline{B(0,1)}$ is weakly compact, hence for each $\varphi\in C([0,1])'$ we have $f\in B$ with $\varphi(f) = \|\varphi\|$. Now look at $\displaystyle\varphi(f) := \int_0^{\frac 12} f(x)dx-\int_{\frac 12}^1f(x)dx$.
A: If you like cannon shooting note that $C[0,1]$ is separable by the (Stone-)Weierstaß theorem. On the other hand, $C[0,1]^*$ is not reflexive as the total variation of $\delta_x-\delta_y$ (which is the norm in $C[0,1]^*$), $\|\delta_x - \delta_y\|=2$ for distinct $x,y\in [0,1]$. This means that $C[0,1]^*$ is not separable as $\{\delta_x\colon x\in [0,1]\}$ is an uncountable discrete subset of $C[0,1]^*$. 
By the Hahn-Banach theorem, dual of a non-separable space is non-separable, hence $C[0,1]^{**}$ cannot be isomorphic to $C[0,1]$ (note that this is stronger that non-reflexivity).
Moral: a separable space with non-separable dual cannot be reflexive.
A: I add another proof. Suppose $C[0,1]$ is reflexive. $C[0,1]$ is separable and thus by reflexivity $C[0,1]*$ is separable. Thus by Helley selection theorem every bounded sequence of continous linear functionals $\varphi_n : C[0,1]^* \rightarrow R$ has a *-weakly convergent subsequence, that is $\varphi_{n_k}(\lambda) \rightarrow \varphi(\lambda)$ for every $\lambda \in C[0,1]*$ Consider the functionals $\lambda_{x}(f) = f(x)$ with $x\in [0,1]$ And the sequence of functions $f_n(x) = x^n$ You get that $J(f_n)$ is a sequence of functional in $C[0,1]**$ and thus you must have, for a subseque, $\lambda_{x}(f_n) = f_n(x) \rightarrow f(x)$ by reflexivity, thus there is a subsequence of $f_n$ that converges pointwise to a continous function. Contradiction
A: Here's another, more elementary proof.
Consider the sequence of functions $f_n=(x\mapsto x^n)\in C[0,1]$. Since it is bounded, by Banach-Alaoglu, it has a limit point $f^{**}\in(C[0,1])^{**}$ (with respect to the weak topology).
Define $f_m^*\in (C[0,1])^*$ by the formula $f_m^*(f)=f(1-1/m)$, while $f^*_{\infty}(f)=f(1)$. Now, notice that we have, for all $m$ that $f^{**}(f_m^*)=0$, while $f^{**}(f_\infty^*)=1$ (because $f_m^*(f_n)\to 0$ as $n\to \infty$ and $f_\infty^*(f_n)$ is identically $1$). Clearly, this is impossible for any $f^{**}\in C[0,1]$, so we must have $f^{**}\in (C[0,1])^{**}\setminus C[0,1]$.
This proof is a little dishonest: to understand the idea behind it, you need a more advanced tool, namely the Riesz-Markov-Kakutani representation theorem. It says that (in particular) the dual of $C[0,1]$ consists exactly of signed Borel measures on $[0,1]$ of bounded variation. It follows immediately that any bounded Borel function on $[0,1]$ represents a unique element of the double dual. The previous two paragraphs just give a concrete example.
