Firstly, note that $\varnothing \subseteq I , \forall I\in O(\mathbb{R})$ where $O(\mathbb{R})$ denotes the collection of all open intervals in $\mathbb{R}$ . Secondly, since the measure $l:O(\mathbb{R}) \rightarrow [0,+\infty)$ is nonnegative, we know that $0 \leq \sum_{i \in \mathbb{N}}l(I_i)$ for all $I_i \in O(\mathbb{R}), i \in \mathbb{N}$.
Now fix $\epsilon > 0$ and set $I_i=(-\epsilon/2^{i+1},\epsilon/2^{i+1}), i=1,\dotsc,n$, which is in $O(\mathbb{R})$. Then
$$0 \leq m^*(\varnothing)=\inf\{\sum_{i=1}^n l(R_i): \varnothing \subseteq \bigcup_{i=1}^n R_i, R_i \in O(\mathbb{R}) \} \leq \sum_{i=1}^n l(I_i) \leq \sum_{i=1}^\infty l(I_i)=\epsilon.$$
Since $\epsilon>0$ is arbitrary, we conclude that $m^*(\varnothing)=0$ for your "finite" outer measure definition. It may be worth mentioning that the countable finite covering has serious implications in some relevant examples. Take $E=\mathbb{Q}\cap [0,1]$ and find $m^*(E)$ in comparison with the countable infinite version $\mu^*(E)$ where $\mu^*(E)=\inf\{\sum_{i=1}^\infty l(R_i): E \subseteq \bigcup_{i=1}^\infty R_i, R_i \in O(\mathbb{R}) \} $, taking into account the denseness of $\mathbb{Q}$ in $[0,1]$.
Extending your question to higher dimensions, if we work with intervals in $O(\mathbb{R^d})$, and then with the product measure $\mu$, we can put instead $I_i=(a_1,b_1)\times \cdots \times (a_d,b_d)$ where $a_k=-\epsilon^{(1/d)}/2^{i/d +1}$ and $b_k=\epsilon^{(1/d)}/2^{i/d +1}, k=1,\dotsc,d$, for some given $\epsilon >0$. Then $\mu(I_i)=(a_1,b_1)\dotsc (a_d,b_d)=(\epsilon^{(1/d)}/2^{i/d})^d=\epsilon/2^{i}$. Therefore, analogously,
$$0\leq m^*(\varnothing)=\inf\{\sum_{i=1}^n \mu(R_i): \varnothing \subseteq \bigcup_{i=1}^n R_i, R_i \in O(\mathbb{R^d}) \} \leq \sum_{i=1}^n \mu(I_i) \leq \sum_{i=1}^\infty \mu(I_i)=\epsilon. $$
From this we have that $m^*(\varnothing)=0$.
[Probably few appreciate the kind of answer here, where instead of answering the poster's question, one poses a series of related easier and harder questions. Let me know if you don't like it and I may stop.]
If there is a problem you can't solve then there is a harder problem you can't solve and probably even an easier problem you can't solve. In this case here are several easier problems which I believe you can solve and which might give you a clue...and a few harder problems to ponder later.
Problem 1. Let $E=(0,1)$, i.e., the open interval with endpoints $0$ and
$1$. Show that for any $\epsilon>0$ there is a finite
pairwise-disjoint collection of measurable sets $\{E_1,E_2,\dots,
E_n\}$ so that each $E_i$ has Lebesgue measure smaller than $\epsilon$
and $\bigcup_{i=1}^n E_i=E$.
Problem 2. Let $E$ be a measurable subset of $(0,1)$, i.e., a subset of the open interval with endpoints $0$ and
$1$. Show that for any $\epsilon>0$ there is a finite
pairwise-disjoint collection of measurable sets $\{E_1,E_2,\dots,
E_n\}$ so that each $E_i$ has Lebesgue measure smaller than $\epsilon$
and $\bigcup_{i=1}^n E_i=E$.
Problem 3. Let $E$ be a measurable set of finite measure. Show that for any $\epsilon>0$ there are numbers
$-\infty<a<b<+\infty$ so that the sets $E\cap (-\infty,a)$ and $E\cap
(b,\infty )$ are measurable sets each of which has Lebesgue measure
smaller than $\epsilon$.
After solving the first three problems the OP's problem should seem quite accessible:
Angelo's Problem. Let $E$ be a measurable subset of finite measure. Show that, for any $\epsilon>0$, there is a finite pairwise-disjoint
collection of measurable sets $\{E_1,E_2,\dots, E_n\}$ so that each
$E_i$ has Lebesgue measure smaller than $\epsilon$ and
$\bigcup_{i=1}^n E_i=E$.
Now let's have more fun.
Harder Problem. Let $E$ be a measurable subset of finite measure equal to $1$ and let $n$ be a positive integer. Show that there is a finite pairwise-disjoint
collection of measurable sets $\{E_1,E_2,\dots, E_n\}$ so that each
$E_i$ has the same Lebesgue measure and
$\bigcup_{i=1}^n E_i=E$.
This problem says that at a birthday party with a one pound, one-dimensional, cake you can split into $n$ equal weight pieces so no-one is shortchanged.
This suggests a much more interesting problem (which I quote from Real Analysis (BBT) Exercise 2:13.8 (Liaponoff's theorem):
Cake Problem: Given a cake with $k$ ingredients (e.g., butter, sugar, chocolate, garlic, etc.) each nonatomic and of unit mass and
mixed together in any "reasonable" way, it is possible to cut the cake
into $k$ pieces such that each of the pieces contains its share of
each of the ingredients.
Best Answer
Let $p\in E$ be arbitrary, $I_1=[p-M,p+M]$. Then $E\subseteq I_1$, hence $m^*(E)\le l(I_1)$.