Completing the proof using strong induction for $E = \bigcap_{n=1}^\infty E_n $
Your inductive step, or maybe your proof as a whole, can perhaps be simplified by one of the following two viewpoints:
1. A dynamical viewpoint
Consider the 10-tupling map $f:[0,1]\to [0,1]$,
$$f:x \mapsto 10x \mod 1,$$
where one multiplies $x$ by $10$ and removes the integer part (another way of writing it: $f(x) = 10x - \lfloor10x \rfloor$).
Who cares?
Why define this function? It makes most sense in digit form:
if $x = 0.d_1d_2d_3\ldots,\ $ $f(x) = 0.d_2d_3d_4\ldots$
That is, $f$ acts by shifting all the digits up by one, forgetting the first digit.
From this alone, you can very quickly verify the following points, and you only need the first.
- If $x\in E$, $f(x)\in E$.
- Moreover, $f: E \to E$ is a 2-to-1 surjection.
- For any $n\in \mathbb N$, $f(E_{n+1}) = E_n$. ($f(E_1) = [0,1]$.)
Regarding the second step of your induction: (EDITED.)
A consequence of property 3.:
$$ y \in \bigcap_{n=1}^{\infty} E_n \implies \ \forall\,n\geq 2,\ y \in E_n \stackrel{\text{3.}}\implies \ \forall\,n\geq 2,\ f(y) \in f(E_n) = E_{n-1} \implies f(y) \in \bigcap_{n=1}^\infty E_n. $$
I.e.,
$$ y \in \bigcap_{n=1}^{\infty} E_n \implies f(y) \in \bigcap_{n=1}^\infty E_n.\tag{1} $$
Now suppose $y = 0.d_1d_2d_3\ldots \in \bigcap_{n=1}^\infty E_n$. The digits of $f(y)$ are given by
$$f(y) = 0.e_1e_2e_3\ldots \qquad\text{where, for every }n,\ e_n = d_{n+1}.$$
By (1), we know $f(y) \in \bigcap_{n=1}^\infty E_n$ and so, by your "base case," its first digit is in $\{4,7\}$.
But this first digit, $e_1 = d_2$, is simply the second digit of $y$, so you're done: $d_2 \in \{4,7\}$.
Hopefully you can see how this can be generalised to $d_{n+1}$ (apply $f$ more than once). No inductive assumption required!
2. Aside: a fractal geometry viewpoint
$E$ is really just like the (middle-thirds) Cantor set. So any argument that works for the Cantor set should work here as well (the Cantor set can be defined in terms of base 3 [ternary], rather than base 10, expansions).
The big result you are trying to prove is a special case of one in fractal geometry.
to set this up, it is simple to see that $E$ is an attractor of the following contractions, $f_4, f_7: [0,1] \to [0,1],$
$$f_4(x) = \frac{x+4}{10},\qquad f_7(x) = \frac{x+7}{10}$$
(in digit form, these are: $f_i:0.d_1d_2d_3\ldots \; \mapsto 0.\,\underline{i}\,d_1d_2d_3),$
where attractor here means: $E$ is a compact subset of $[0,1]$ ($[0,1]$ being the domain) such that $E = f_4(E) \sqcup f_7(E)$.
From fractal geometry, it is known that attractors are 1) unique and 2) satisfy the following formula.
$$E = \bigcap_{n= 1}^\infty \;\; \bigcup_{(i_1,\ldots,i_n)\in \{4,7\}^n} \underbrace{f_{i_1}\circ f_{i_2}\circ f_{i_3}\circ \cdots \circ f_{i_n}[0,1]}_{\text{first $n$ digits are }i_1,\ i_2,\ldots i_n} $$
Furthermore, $$ E_n = \bigcup_{(i_1,\ldots,i_n)\in \{4,7\}^n} f_{i_1}\circ f_{i_2}\circ f_{i_3}\circ \cdots \circ f_{i_n}[0,1],$$
where this last equality follows by your definition of $E_n$ E.g.,
$$E_1 = f_4[0,1] \cup f_7[0,1]$$ $$E_2 = f_4\circ f_4[0,1] \cup f_4\circ f_7[0,1] \cup f_7\circ f_4[0,1] \cup f_7\circ f_7[0,1] $$
and so on.
In other words, this formula (or its proof) gives you another altenrative proof/viewpoint.
Other key words which apply: iterated function system, self-similar sets.