We may take $\epsilon=1$ as if $\bigcup E_i$ has eccentricity constants $1-\delta$ and $1+\delta$, then $\bigcup rE_i$ has eccentricity constants $r(1\pm\delta)$. The thing that you are really trying to control is the ratio of the eccentricity constants. Here is a construction with boundedthe ratio of the eccentricity constants arbitrarily close to 2. I don’t see how you could get themthe ratio to be arbitrarily close to 1 though.
Take a countable dense collection of functions. Then choose a maximal sub-collection of functions that are at least 1 unit apart and let $\eta>0$. Any continuous function is now within 1.1$1+\eta$ of this collection- otherwise it can be approximated by something in the dense collection, contradicting the maximality.
Now call your functions $(f_n)$. Define $S_n=\{g:\|g-f_n\|\le 1.1\inf_m\{\|g-f_m\|\}\}$$S_n=\{g:\|g-f_n\|\le (1+\eta)\inf_m\{\|g-f_m\|\}\}$. Finally, take $B_n=S_n\setminus\bigcup_{j<n}S_j$.
If $\|g-f_n\|<\frac 1{2+\eta}$, then since the $f_j$'s are 1-separated, $\|g-f_m\|>\frac{1+\eta}{2+\eta}$ for any $m\ne n$. Hence $g\in S_n$, so that we have shown $B(f_n,\frac 1{2+\eta})\subset B_n$ On the other hand, if $\|g-f_n\|\ge(1+\eta)^2$, then we have already shown that there exists $m$ with $\|g-f_m\|<1+\eta$, so that $g\not\in S_n$. It follows that $B(f_n,\frac 1{2+\eta})\subset B_n\subset B(f_n,1+\eta)$. Since $\eta$ can be chosen arbitrarily close to 0, we can get a ratio of eccentricity constants as close to 2 as desired.