prove that $$\lambda^{*}(A) = \inf \Big\{\sum_{k=1}^{\infty} \lambda(I_k) \Big| A \subset \bigcup_{k=1}^{\infty} I_k \Big\}$$
It suffices to show that
$$\forall \epsilon > 0 , \exists (I_k) \text{ and }A \subset \bigcup_{k=1}^{\infty} I_k \text{ s.t } \sum_{k=1}^{\infty} \lambda (I_k) - \epsilon < \lambda^{*}(A) $$
note that outer measure is defined as follows
$$\lambda^{*}(A) = \inf_{A \subset G : \text{G is open}} \lambda(G)$$
so given $\epsilon,$ there exists $A \subset G$ s.t $\lambda(G) - \epsilon < \lambda^{*}(A)$
now use following property
every nonempty open subset of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping axis rectangles, i.e.,$G = \bigcup_{k=1}^{\infty} I_k$ where $I_k$ are axis aligned rectangles.
$\sum_{k=1}^{\infty} \lambda(I_k) - \epsilon = \lambda(G) - \epsilon < \lambda^*(A)$ so, we are done.