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
If we replace $I_k$ with $\bar I_k$ then $l(I_k)=l(\bar I_k).$ On the other hand if $ G=\{J_k :k\in N\}$ is a set of closed bounded intervals with $A\subset \cup G,$ then for any $e>0$ we can cover each $J_k$ with a bounded open interval $I_k$ with $l(I_k)<e 2^{-k}+l(J_k)$. So $m*=m**$. It is a matter of convenience. For example with the usual def'ns of outer measure $m^o$ and inner measure $m^i$ it is fairly obvious that if $A$ is a bounded interval and $B\subset A$ then $m^i(B)+m^o(A\backslash B)=l(A)$.