$c_0(C[0,1])$ is the $c_0$-direct sum of countably many $C[0,1]$.How to prove
$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$.
Here,Banach-space isomorphism means a bounded invertible operator from $C[0,1]$ onto $c_0(C[0,1])$.
$c_0(C[0,1])$ is the $c_0$-direct sum of countably many $C[0,1]$.How to prove
$C[0,1]$ is Banach-space isomorphic to $c_0(C[0,1])$.
Here,Banach-space isomorphism means a bounded invertible operator from $C[0,1]$ onto $c_0(C[0,1])$.
There is a useful simple
Lemma. If $X\sim X\oplus X$, $Y\sim Y\oplus Y$, and each of $X,Y$ is isomorphic to a complemented subspace of another, then $X\sim Y$.
Proof. We have $X\sim Y\oplus A$, then $X\sim (Y\oplus Y)\oplus A=Y\oplus(Y\oplus A)=Y\oplus X$, analogously $Y\sim X\oplus Y$.
Now let $X=C([0,1])$, $Y=c_0(X)$. Property $Y\sim Y\oplus Y$ is clear, property $X\sim X\oplus X$ follows from $X\oplus X=C([0,1]\times \{0,1\})$ and Milyutin theorem. $X$ is clearly complemented in $Y$. At last, $Y$ is the space of functions on the compact space on the plane $K=\cup \{\frac1n \times [0,\frac1n]\}\cup (0,0)$, which are equal to 0 at $(0,0)$. Again by Milyutin theorem we see that $X\sim C(K)$, hence $Y$ is isomorphic to a hyperplane in the space isomorphic to $X$. Hyperplane is of course complemented.
This is maybe not a very good proof, since your claim may be used in the proof of Milyutin theorem (I do not remember). But if so, study the proof, it should contain this claim.
If you mean $c_0(X)=\lbrace (x_n)_{n\in\mathbb N} \in X^{\mathbb N}: \|x_n\|_X\to 0\rbrace$ with norm $\|(x_n)_n\|=\sup\lbrace \|x_n\|_X: n\in\mathbb N\rbrace$, then $c_0(C[0,1])$ and $C[0,1]$ are isomorphic (as explained by Fedor) but not isometrically isomorphic: The unit ball of $C[0,1]$ has extreme points but that of $c_0(C[0,1])$ does not.