$E$ is in fact closed. Take $x = \overline{0.a_1a_2\cdots} \in E^\complement$, where the $n$-th decimal place $a_n \in \{0,\dots,9\}$ for all $n \in \Bbb{N}$. Since $x \notin E$, there exists $m \in \Bbb{N}$ such that $a_m \notin \{4, 7\}$. Let's choose the least possible value of $m$, and choose $r$ sufficiently small so that $B_r(x) \subseteq E^\complement$ by defining $r = \color{red}{\frac{1}{100}\min}\{|x - c_i| \mid i \in \{1,\dots,4\}\}$. Observe that $E$ is "locally bounded" by the intervals $[c_1,c_2]$ and $[c_3,c_4]$.
$$\bbox[36px, yellow, border: 2px solid red]{\begin{array}{rrrrrrrrrr}
\rlap{\style{display: inline-block; transform: scale(10,1)}{-}} {\Large [} & {\large \bullet} & \llap{\style{display: inline-block; transform: scale(10,1)}{-}} {\Large ]} & {\style{display: inline-block; transform: scale(10,1)}{-}} & {\Large |} {\style{display: inline-block; transform: scale(10,1)}{-}} & {\style{display: inline-block; transform: scale(10,1)}{-}} {\Large |} & {\style{display: inline-block; transform: scale(10,1)}{-}} & \rlap{\style{display: inline-block; transform: scale(10,1)}{-}} {\Large [} & {\large \bullet} & \llap{\style{display: inline-block; transform: scale(10,1)}{-}} {\Large ]} \\
c_1 & \begin{matrix}\uparrow \\ E\end{matrix} & c_2 & \rlap{\small\overline{0.a_1a_2\cdots a_{\color{red}{m-1}}5/6}} & & & & c_3 & \begin{matrix}\uparrow \\ E\end{matrix} & c_4
\end{array}}\\
\text{Figure 1: range of possible values of numbers in $E$}$$
- $c_1 = \overline{0.a_1a_2\cdots a_{\color{red}{m-1}} \color{red}{4}4}$
- $c_2 = \overline{0.a_1a_2\cdots a_{\color{red}{m-1}} \color{red}{4}8}$
- $c_3 = \overline{0.a_1a_2\cdots a_{\color{red}{m-1}} \color{red}{7}4}$
- $c_4 = \overline{0.a_1a_2\cdots a_{\color{red}{m-1}} \color{red}{7}8}$
Recall that $x = \overline{0.a_1a_2\cdots a_{\color{red}{m-1}} a_m a_{m+1} a_{m+2} \cdots}$, so $$r \le \frac{44 \cdot 10^{-(m+1)}}{100} < \frac{50 \cdot 10^{-(m+1)}}{100} = \frac{5}{10^{m+2}}.$$ When we add/minus $r$ to/from $x$, the new number $x \pm r$ "won't differ too much from $x$". Consider $x - r$ ( or $x + r$). Either one of the following cases occurs:
- the $m$-th decimal place is still $a_m \notin \{4,7\}$, so $x - r \notin E$.
- the $m$-th decimal place is changed by $1$ (digit $0 \to 9$ also counts). A necessary condition for this to happen is that $a_{m+1} \in \{0,9\}$. (To see this, imagine examples like $0.95 + 0.05 = 1$.) In this case, the $(m + 1)$-th decimal place of $x-r$ would be either $0$ or $9$, so $x - r \notin E$.
Thus, any number in $B_r(x)$ can't belong to $E$. Therefore, $E^\complement$ is open and $E$ is closed. Since $E$ is bounded, Heine–Borel Theorem tells us that $E$ is compact.
Your mistake is the wrong deduction from "$B_r(x)$ must contain a number at least one of whose digits contains $4$ or $7$" to "$E^\complement$ is not open". To be a member of $E$, the decimal $y \in B_r(x)$ should consist merely in digits $4$ and $7$, not just "at least one" digit.
Remarks: The choice of $r$ is a bit tricky and the verification is a bit tedious, but it's still doable.
$E$ is NOT a perfect set. The idea that $$0.4\overset{\bullet}{7}, 0.44\overset{\bullet}{7}, 0.444\overset{\bullet}{7}, \cdots \to 0.\overset{\bullet}{4}$$ only holds for a member of $E$ which is a non-terminating decimal. In that case, each term $0.4\dots4\overset{\bullet}{7}$ on LHS contains digit $7$, so it's different from RHS $0.\overset{\bullet}{4}$, and thus it lies in a deleted neighbourhood of $0.\overset{\bullet}{4}$. To finish the argument, just repeat this idea for any $x \in E$ which is a non-terminating decimal.
For any $x = \overline{0.a_1a_2\cdots} \in E$ which is a non-terminating decimal, I'm going to construct a sequence $(y_n)_n$ in $E$ so that $\lim\limits_n y_n = x$. For all $n \in \Bbb{N}$, define $y_n = \overline{0.b_1b_2\cdots b_{m-1} b_m b_{m+1}}$, where $$b_m = \begin{cases} a_m \quad &\text{ if } m \le n \\ \max(\{4,7\}\setminus\{a_m\}) & \text{ if } m > n \end{cases}.$$ $\require{HTML}$
$$
\bbox[yellow, 5px, border: 2px solid red]{
\begin{array}{rrrrrrrrrr}
x & = & 0. & a_1 & a_2 & \cdots & a_{n-1} & a_n & a_{n+1} & \cdots \\
\style{display: inline-block; transform: rotate(90deg)}{\ne} & & & \style{display: inline-block; transform: rotate(90deg)}{=} & \style{display: inline-block; transform: rotate(90deg)}{=} & \style{display: inline-block; transform: rotate(90deg)}{=} & \style{display: inline-block; transform: rotate(90deg)}{=} & \style{display: inline-block; transform: rotate(90deg)}{=} & \style{display: inline-block; transform: rotate(90deg)}{\ne} & \style{display: inline-block; transform: rotate(90deg)}{\ne} \\
y_n & = & 0. & b_1 & b_2 & \cdots & b_{n-1} & b_n & b_{n+1} & \cdots
\end{array}} \\
\text{Figure 2: construction of approaching sequence $(y_n)_n$ in $E'$ from $x \in E$}
$$
It's clear that for all $n \in \Bbb{N}$, $y_n \ne x$ and $y_n \in E$, but $\lim\limits_n y_n = x$, so $x \in E'$.
If $x \in E$ is a terminating decimal, as Josie's comment below suggests, just take a "suitably small" $r$ to that $B_r(x)\setminus\{x\} \subset E^\complement$, say to take $x = 0.4$ and $r = 0.02$, so that the deleted neighbourhood becomes $(0.38, 0.4) \cup (0.4, 0.42)$, which contains no member of $E$.
Best Answer
The number $0.7$ does not belong to $E$, since $0.7=0.70000\ldots$. Elements of $W$ are, for instance $0.77777\ldots$, $0.7474474447\ldots$ and so on.