To construct this covering it is enough to, for a fixed $\varepsilon$, give a collection of disjoint balls in $C[0,1]$ of radius $\varepsilon$ which is dense in $C[0,1]$. Indeed, then for $\delta$ as small as you want you can take the sets $E_i$ to be the balls and adjoin the points outside them to one ball at distance $<\delta$ of them (there are countable balls so you can do this in such a way that the resulting sets are Borel), obtaining a covering such that each point is in just one set, and with eccentricity constants $\varepsilon,\varepsilon+\delta$.
To build this collection of balls we first consider a sequence of distinct points $0,1,a_3,a_4,\dots$ dense in $[0,1]$. Now for $n\geq2$, let $A_n$ be the set of functions which take some arbitrary values in the points $a_1,a_2,\dots,a_n$ and are defined by linear interpolation in the rest of the interval. Then $A_n$ is isometric to $\mathbb{R}^n$ in the square metric, with the coordinates given by the values of the function in $a_1,\dots,a_n$. Moreover, in this coordinates, the inclusion of $A_n$ in $A_{n+1}$ is given by a linear inclusion of $\mathbb{R^n}$ into $\mathbb{R}^{n+1}$ which preserves the first $n$ coordinates.
Now we can consider a tiling of $A_2$ by balls of radius $\varepsilon$ (in the square metric of $\mathbb{R}^2$, squares of side $2\varepsilon$), and we call $C_2$ the collection of their centers. As the inclusion $A_2\to A_3$ is linear and preserves the first $2$ coordinates, we can extend this tiling to a tiling of $A_3$ by cubes of radius $\varepsilon$. More precisely, the set $C_3$ of centers of cubes in $A_3$ will be given in coordinates by $\{p+k(0,0,\dots,0,2\varepsilon);p\in C_2,k\in\mathbb{Z}\}$. Concretely, that implies $C_2\subseteq C_3$. This way we can extend this tilings of $A_n$ to tilings of $A_{n+1}$ by balls (hypercubes) of radius $\varepsilon$, with centers in a set $C_n$. Note that by construction, the points of the $C_n$ are at distance $\geq 2\varepsilon$ from each other.
So, the collection of balls of radius $\varepsilon$ and center in $\cup_{i=1}^\infty C_n$ will be disjoint, and it is dense in $\cup_{n=2}^\infty A_n$. As $\cup_{n=2}^\infty A_n$ is dense in $C[0,1]$, the collection of balls is dense in $C[0,1]$ too. So now we can just use the construction of the first paragraph to get the cover we want.