TITLE: The outer measure of a closed interval is its length without Heine-Borel Theorem.
QUESTION [5 upvotes]: I tried to prove that the outer measure of a closed interval is its length without Heine-Borel Theorem. Proving the first part  $m^\star ([a,b])\leq a-b$ is easy and does not require Heine-Borel Theorem.
Proving the other part  is summarized as follows:
(1)-Clearly we have $[a,b]\subset(a-\epsilon ,b+\epsilon ) $ for any scalar $\epsilon>0$.
(2)- $a$ and $b$ are interior points of  any open coverng since $[a,b]$ $\subset$ $\cup_{k=1}^\infty I_k$.
(3)- The covering $\cup_{k=1}^\infty I_k$ is open. Hence there exists $\epsilon_1>0$ and $\epsilon_2>0$ such that  the intervals $(a-\epsilon_1,a+\epsilon_1)$ and $(b-\epsilon_2,b+\epsilon_2)\subset  \cup_{k=1}^\infty I_k$.
(4)- Steps (1), (2), and (3) imply that    there exists $\epsilon=min\{\epsilon_1,\epsilon_2\}$ such that $(a-\epsilon ,b+\epsilon )\subset  \cup_{k=1}^\infty I_k$ for every open covering of $[a,b]$, i.e. for any open covering $\cup_{k=1}^\infty I_k$ we have an open cover (not subcover) that contains $[a,b]$. In other words, for any open cover, there exists $\epsilon>0$ such that $(a-\epsilon ,b+\epsilon )$ is an open cover of $[a,b]$ with the following property:
$$[a,b]\subset(a-\epsilon ,b+\epsilon ) \subset  \cup_{k=1}^\infty I_k$$
I know that the interval $(a-\epsilon ,b+\epsilon )$ is not a subcover but it is an open cover that is tighter than the open cover itself. We can take the infimum over $\epsilon$ which is the same as taking the infimum overall covering.
$\textbf{EDIT: We have}$ $b-a\leq l((a-\epsilon,b-\epsilon)\leq\sum_kl(I_k))$, $\textbf{which implies:}$
$$b-a= \inf_{\epsilon>0}\{ b -a+2\epsilon \}\leq \inf\bigg\{\sum_{k=0}^\infty l(I_k): [a,b]\subset\cup_{k=0}^\infty I_k\bigg\}= m^\star ([a,b])$$
which completes the proof.
Note that I did not use the subcovering property. I used the fact that every open covering of a closed interval contains an open interval (which is not a subcover) that covers the closed set.
Is there any flaw in my proof?

REPLY [2 votes]: When you write [I've made a small correction]
\begin{equation}
\label{4385212:eq:1}\tag{A}
\inf_{\epsilon>0} (b - a + 2\epsilon) \leqslant
\inf\left\{\sum_{k=0}^\infty l(I_k) : [a, b] \subset
\bigcup_{k=0}^\infty I_k\right\}
\end{equation}
you appear to be appealing to an unstated lemma to the effect that
if $J$ is an open interval, and $(I_k)_{k\geq0}$ is an infinite
sequence of open intervals, and
$$
J \subseteq \bigcup_{k=0}^\infty I_k,
$$
then
$$
l(J) \leqslant \sum_{k=0}^\infty l(I_k).
$$
Given such a lemma, it would be valid to infer \eqref{4385212:eq:1}
from your previous assertion that for all open $(I_k)_{k\geq0}$ such
that $[a, b] \subset \bigcup_{k=0}^\infty I_k,$ there exists
$\epsilon > 0$ such that
$$
(a - \epsilon, b + \epsilon) \subset \bigcup_{k=1}^\infty I_k,
$$
and I think the proof would be OK - but do you have a proof of the lemma?