You have the right idea, but the notation could be improved in your second step. If $(R_j)_j$ is an enumeration of a countable covering of $E$ with rectangles, I wouldn't bother separating the rational ones from the ones that are not: Fix $\varepsilon > 0$. For every $j$, $R_j$ is contained in a slightly bigger rational rectangle $R^{**}_j$, say with $|R^{**}_j| \leq |R_j| + \varepsilon/2^j$. Thus
$$\sum_j |R_j| \geq \sum_j |R^{**}_j| - \varepsilon \geq \inf \sum_j |Q_j| - \varepsilon,$$
where the infimum on the right runs over all rational rectangles covering $E$. Taking the infimum over all $(R_j)_j$ and then letting $\varepsilon \to 0$, we get the result together with step 1.
In the context of $\mathbb R$, explicitly naming the sequence of open intervals is not really necessary. Since every open subset of $\mathbb R$ is the union of such a countable sequence, you can restate the definition of outer measure as follows, first given an arbitrary open set $U = \cup \{ I_k: k \in \mathbb{N} \} \subset \mathbb R$ define,
$$ m^\ast(U) = \sum_{k=0}^\infty \ell(I_k)$$
then, extend the definition of $m^\ast$ to arbitrary subsets $E \subset \mathbb R$ by asserting $m^\ast(E) = $ $\inf \{ m^\ast(U): E \subset U \text{ and } U \text{ is open } \}$.
What does this simplification/restatement accomplish? It enables a cleaner analysis of the process by which the open sets $\mathcal{U} = \{ U \subset \mathbb{R}: U \text{ is open and } E\subset U\}$ are used to find the outer measure of $E$.
Recalling that the infimum $M = \inf \{ m^\ast(U): U \in \mathcal{U}\} = m^\ast(E)$ must satisfy,
For every $\delta > 0$, there exist some $U \in \mathcal{U}$, such that, $ M \le m^\ast(U) < M - \delta$
we can choose/find/fix a sequence of open sets $U_0, \ldots, U_n, \ldots \in \mathcal{U}$ such that, for each $n\ge 0$,
$$ M \le m^\ast(U_n) < M - \frac{1}{2^n}. $$
Letting $V_n = U_0 \cap \ldots \cap U_n$, we get, $V_n \in \mathcal{U}$
(since $E \subset U_0 \cap \ldots \cap U_n = V_n$ and $V_n$ is open.) Moreover, applying the definition of $m^\ast$ for open sets, we have $V_{n+1} \subset V_n \implies$ $m^\ast(V_{n+1}) \le m^\ast(V_n)$,and
$$ M \le m^\ast(V_n) \le \min \{ m^\ast(U_0), \ldots, m^\ast(U_n) \} < M - \frac{1}{2^n}.$$
Putting this all together, we've extracted a descending sequence $V_n$ of open sets containing $E$, whose measures converge to the value of the outer measure $M = m^\ast(E)$ of $E$.
You can think of the sets $V_k$ as approximating $E$ by shrinking and progressively discarding non-essential segments. Moreover, capturing that processing is the whole idea behind how the outer measure is defined.
Best Answer
Take the following set of rectangles:
$$Q_k=[k,k+1]\times\left[-\frac{\epsilon}{2^{|k|}},\frac{\epsilon}{2^{|k|}}\right] \;\;\text{ for each }\;\; k\in \mathbb{Z}$$
$$m_*(E)\leq \sum\limits_{k\in \mathbb{Z}}|Q_k|= 2\epsilon + 2\sum_{k\geq 1}\frac{\epsilon}{2^{k-1}}=6\epsilon$$
And you can take $\epsilon>0$ arbitrarily small, that is to say, that $m_*(E)\leq 0$.
As you can see, the trick is to make each covering's measure finite and dependant of a free parameter, in this case $\epsilon$.