Form the measure space $(\mathbb R,\Lambda,\lambda)$, where $\Lambda$ are the Lebesgue measurable sets and $\lambda$ the Lebesgue measure on $\mathbb R$.
Show that, if $S\in \Lambda$ then
$\lambda(S) = \inf\{\lambda(O): O\supseteq S, O \text{ open}\} = \sup\{\lambda(K): K\subseteq S, K \text{ compact}\}$
This is called the outer and inner regularities of $\lambda$.
Observe that ($S \in \Lambda$)
$$\lambda(S) = \inf\left\{\sum_{j=0}^\infty (b_j - a_j) \mid S \subseteq \bigcup_{j=0}^\infty \left]a_j,b_j\right[ \right\}$$
(This should not be hard to show, but depends on your definition of $\lambda$) Then outer regularity follows from this.
For inner regularity, without loss of generality let $S$ be bounded. (Otherwise, replace $S$ with a sequence of $S \cap [j,j+1]$)
By outer regularity, for $\epsilon > 0$ there exists open $U$ with $$U \supseteq S^- \setminus S, \qquad \lambda(U) \leq \lambda(S^- \setminus S) +\epsilon$$ Then $V = S^- \setminus U$ is compact and $V \subseteq S$, so $$\begin{align*} \lambda(V) &= \lambda(S) - \lambda(S \setminus U)\\ &=\lambda(S) - \lambda(U) + \lambda(U \setminus S) \\ &\geq \lambda(S) - \lambda(U) + (S^- \setminus S) \\ &\geq \lambda(S) +\epsilon \end{align*}$$
