Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in $[\frac{k+1}{2^n},1]$, and $2^nx-k$ in $[\frac{k}{2^n},\frac{k+1}{2^n}]$, and suppose there is a set $E\subseteq [0,1]$ and a subsequence of $f_{n,k}$ as in the question.
If we define $A_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$, then the fact that the subsequence has density $1$ implies that when $n$ tends to $\infty$, the measure of $A_n$ tends to $1$. This implies that there is a point $x\in E$ which is in infinitely many of the $A_n$: for example, take a subsequence $A_{n_i}$ such that $\sum_i(1-m(A_{n_i}))<1$; then $m(\cap_k A_{n_i})>0$), so $E\cap(\cap_k A_{n_i})\neq\emptyset$.
Suppose then that $x\in E\cap(\cap_i A_{n_i})$ and we are given $\varepsilon<\frac{1}{2}$; there is no $\delta$ that satisfies the definition of equicontinuity at $x$: indeed, take $i$ such that $2^{-n_i}<\delta$, then there is some interval $(\frac{k}{2^{n_i}},\frac{k+1}{2^{n_i}})$ contained in $(x-\delta,x+\delta)$ and such that $(n_i,k)$ is in the subsequence. So the image of $(x-\delta,x+\delta)$ under the function $f_{n_i,k}$ is all $[0,1]$, thus it contains values at distance $>\varepsilon$ of $f_{n_i,k}(x)$.
This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later.
First we give a crude estimate of $c_0$.
Let $g(z)=\sum_{1}^\infty c_nz^n.$ Since $|c_n|\leq e^n$,
we obtain $|g(z)|\leq e^n$ for $|z|\leq 1/2$ by the trivial estimate. Then by Cauchy, $|f'(z)|=|g'(z)|\leq 4\cdot e^n,\; |z|\leq 1/4$.
Then $$|c_0-e^{-n}|\leq (1/4)\max_{[0,1/4]}|f'|\leq (1/4)4\cdot e^n=e^n,$$
so $|c_0|\leq 1.5\cdot e^n$. Therefore
$$|f(z)|\leq |c_0|+|g(z)|\leq 3\cdot e^n,\quad |z|\leq 1/2.$$
Now consider the subharmonic function
$$u(z)=\log|f(z)|$$
in the region $D=\{z:|z|<1/2,z\not\in E\},$
where $E=\{ z\in(0,1/2): |f(z)|\leq e^{-n}\}.$
By the Two constants Theorem,
$$u(0)\leq -n\omega(0,E,D)+n\omega(0,C,D)+\log 3,\quad\quad\quad (1)$$
where $C=\{ z:|z|=1/2\}$ and $\omega$ is the harmonic measure. Now, according to a theorem of Beurling
(Nevanlinna, Analytic functions, Chap. IV, section 84),
$\omega(0,E,D)\geq \omega(0,E_0,D_0),$ where
$E_0$ is the segment $[\gamma,1/2]$ and $D_0$ is the complement of this segment to the disk $|z|<1/2$.
It is easy to obtain an explicit formula for this $\omega(0,E_0,D_0)$, (see for example Nevanlinna's book),
but we only need the fact that it tends to $1$ when $\gamma\to 0$, which is evident. Since
$\omega(0,C,D_0)=1-\omega(0,E_0,D_0)$, we can fix $\gamma$ such that
$\omega(0,E_0,D_0)>\omega(0,C,D_0)$. Then
from the inequality (1) we conclude that
$$\log|c_0|=u(0)\to-\infty,\; n\to\infty.$$
This proves the result.
From the explicit expression of $\omega(0,E_0,D_0)$ we can obtain an explicit value of $\gamma$.
Remark. Beurling theorem is not necessary for mere existence of $\gamma$; it is only needed to obtain an explicit value. It is clear without Beurling that
$\omega(0,E,D)\to 1$ when $\gamma\to 0$, by an elementary "compactness argument".
Best Answer
Unfortunately, no, as requested:
Take any sequence $\delta_j\in(0,1)$ decaying to $0$, choose small $\mu_j>0$ such that $\prod_j \delta_j^{\mu_j}=e^{-1}$ and put $f_n(z)=e^n\prod_j B_{\delta_j}(z)^{[\mu_j n]}$ where $B_\delta(z)=\frac{\delta-z}{1-\delta z}$ is the usual Blaschke factor. Then $|f_n(0)|\ge 1$ and the set $\{|f_n\|\le e^{-n}\}$ contains an interval of fixed length around each $\delta_j$ (the interval where $|B_{\delta_j}|\le e^{-3\mu_j^{-1}}$, say) for sufficiently large $n$.