I’m going through the construction of Lebesgue measure, which relies on this result. In the following, the $I$ are all intervals defined on $[0,1]$ and $\mathbb{P}(I)$ denotes the length of the interval.
Lemma 1
Let $I_1, I_2, \cdots, I_n$ be a finite collection of intervals whose union contain an interval $I$, then $$\sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I).$$
Proof:
For any $1\leq j \leq n$, let $a_j$ be the left end point of $I_j$ and $b_j$ be the right end point of $I_j$. The sets $\{a_j\}$ and $\{b_j\}$ are finite and thus can be ordered. Let $\{a_{l_i}\}$ and $\{b_{k_i}\}$ be the ordered sets of these end points (from least to greatest).
For the interval $I$ we have $$\bigcup_{j=1}^n I_j \supseteq I.$$
Letting $a$ and $b$ denote the left and right end points of $I$, respectively, without loss of generality we assume $$a_{l_1}\leq a\leq b_{l_1}, $$ $$a_{k_n}\leq b\leq b_{k_n}. $$
The above simply says that $a$ and $b$ are in the intervals with smallest left end point and largest right end point, respectively (this could of course be the same interval). Using the ordering of end points, we can pair them up, from smallest to largest, in the sum of interval lengths:
$$\sum_{j=1}^{n}(b_j - a_j)= \sum_{i=1}^{n}(b_{j_i} - a_{j_i})= \Big[\sum_{i =2}^{n-1}(b_{j_i} - a_{j_i}) +(b_{j_1} -a_{j_n})\Big]+(b_{j_n} -a_{j_1}) .$$
In the last expression, we have removed the smallest left end point and largest right end point from the sum. Since we assume some part of the interval $I$ is in every $I_n$, the remaining end points of intervals must overlap, which means, for any $i$, $$b_{j_{i-1}} \geq a_{j_i}.$$ Thus, we can reorder again to show that
$$\sum_{i =2}^{n-1}(b_{j_i} - a_{j_i}) + (b_{j_1}-a_{j_n})= \sum_{i=2}^{n} (b_{j_{i-1}}- a_{j_i}) \geq0.$$
Returning to our original sum, it follows that
$$\sum_{j=1}^n (b_j - a_j) \geq b_{k_n} - a_{l_1} \geq b-a,$$
which implies $$ \sum_{j=1}^{n} \mathbb{P}(I_j) \geq \mathbb{P}(I).$$ $ \square$
Lemma 2:
Let $I_1, I_2,\cdots$ be a countable collection of open intervals, whos union contains a closed interval, then $$\sum_{j\geq 1}\mathbb{P}(I_j) \geq \mathbb{P}(I).$$
Proof:
By the Heine-Borel theorem, for any countable collection of open sets $\{O_n\}$ and some closed set $ \bigcup_{n\geq1} O_n \supseteq O$, there exists a finite sub cover $$O_{a_1} \cup O_{a_2} \cup \cdots \cup O_{a_k} \supseteq O.$$
Thus $$ \bigcup_{j=1}^\infty I_j \supseteq I. \implies \bigcup_{a=1}^k I_{j_a} \supseteq I$$ and the result follows from lemma 1.
$\square$
Lemma 3:
For any countable collection of intervals $\{I_j\}$, whose union contains an interval $I$, $$\sum_{j \geq 1} \mathbb{P} (I_j) \geq \mathbb{P}(I).$$
Proof:
Choose some $\epsilon >0$ and extend the left and right end points of the $I_j$ by $\epsilon 2^{-j}$ to form an open set $$I_{j_\epsilon}=\left(a_j -\epsilon 2^{-j} , b_j + \epsilon 2^{-j} \right).$$
Next, extend the end points of $I$ by $\epsilon$ to make the closed set $$I_\epsilon =[a- \epsilon , b + \epsilon].$$
Since $\bigcup_{j \geq 1} I_j \supseteq I,$ there exists some $a_k$ and $b_l$ such that
$$a_k <a \leq b < b_l.$$
Further, since $k$ and $l$ are both $\geq 1$, we can choose $\epsilon$ small enough that
$$a_k - \epsilon 2^{-k}< a_k<a - \epsilon \quad, \quad b+\epsilon < b_l < b_l + \epsilon 2^{-l}.$$
Thus, we have constructed a countable collection of open intervals $\{I_{j_\epsilon}\}$, whose union contains the closed interval $I_\epsilon$. By lemma 2:
$$\sum_{j\geq1} \mathbb{P}(I_{j_\epsilon}) \geq \mathbb{P}(I_\epsilon)$$
$$\implies \sum_{j \geq 1} \Big \{ \big( b_j + \epsilon 2^{-j} \big) - \big(a_j - \epsilon 2^{-j} \big)\Big\} \geq( b + \epsilon )- (a - \epsilon).$$
$$\implies \sum_{j \geq 1} ( b_j - a_j) + \epsilon \sum_{j \geq 1} \big (2^{1-j} \big) \geq (b-a) + 2\epsilon.$$
$$ \implies \sum_{j \geq 1} ( b_j - a_j) + 2\epsilon \geq (b-a) + 2\epsilon.$$
$$\implies \sum_{j\geq 1} \mathbb{P}(I_j)=\sum_{j \geq 1} ( b_j - a_j) \geq (b-a) = \mathbb{P}(I).$$ $ \square$
