The space $C^0([0;1],\mathbb{R})$ of all continuous, real-valued functions on $[0,1]$ is not reflexive. How does one prove that $C^0([0;1],\mathbb{R})$ equipped with the supremum norm is not reflexive?

I don't understand how to show that the $J$ mapping is not surjective.
As we have seen in a previous question, if $E$ is a reflexive Banach space then each linear continuous functional attains its norm. So, in order to show that $E:=C^0([0,1],\Bbb R)$ endowed with the supremum norm is not reflexive, it is enough to find a linear functional which is not norm attaining. We can define
$$x'(f):=\int_0^{1/2}f(t)\mathrm dt-\int_{1/2}^1f(t)\mathrm dt;$$
$x'$ is linear and $|x'(f)|\leqslant \lVert f\rVert_{\infty}$, hence $x'$ is continuous and its norm is $\leqslant 1$.
To see that the norm is indeed $1$, for $n$ integer, consider a function $f_n$ which is $1$ on $[0,1/2-1/n)$ and $-1$ on $(1/2+1/n,1)$, and linear on $(1/2-1/n,1/2+1/n)$. We can see that $\lVert f_n\rVert=1$ and $x'(f_n)=1-2/n$.
Now, we have to show that we cannot find $f\in E$ such that $x'(f)=1$ and $\lVert f\rVert=1$. Let $f$ be continuous of norm $1$. We have to show that $x'(f)\neq 1$, for a fixed $\varepsilon&gt;0$ we can find $\delta&gt;0$ such that $|f(t)-f(1/2)|\leqslant \varepsilon$ whenever $|t-1/2|\leqslant \delta$. Since
$$\tag{*}  x'(f)=\int_0^{1/2-\delta}f(t)\mathrm dt+\int_{1/2-\delta}^{1/2+\delta}f(t)\mathrm dt -\int_{1/2+\delta}^1f(t)\mathrm dt,$$
we can assume that $f(x)=1$ on $[0,1/2-\delta]$ and $f(x)=-1$ on $[1/2+\delta,1]$, otherwise it is clear that $x'(f)\neq 1$.
By (*), it follows that $|x'(f)|\leqslant 1-2\delta+2\delta\varepsilon+\delta| f(1/2)|$. Taking $\varepsilon\lt 1-|f(1/2)|$, we get that $|x'(f)|\lt 1$.
you could also use the following facts:


$BV[a,b]$ is not separable, 
$C[a,b]$ is separable, 
if $X$ is a Banach Space and $X^*$ is its dual and is separable, then $X$ is separable. 


Since $BV[a,b]$ is the dual of $C[a,b]$, reflexivity would imply $BV[a,b]$ is separable.