Why is the outer measure of the set of irrational numbers in the interval [0,1] equal to 1?

The rational numbers has measure zero, so $\mathbb{Q}\in \mathcal{M}(\lambda^*)$. Then

\begin{align}\,1=\lambda^*([0,1])=\lambda^*([0,1]\cap\mathbb{Q})+\lambda^*([0,1]\setminus \mathbb{Q})=0+\lambda^*([0,1]\setminus \mathbb{Q})\end{align}

i.e., $1=\lambda^*([0,1]\setminus \mathbb{Q})$. $~~~~~~~$

What you know is that $\sum_k l(I_k) \le m^*(A) + \epsilon$ for some sequence of intervals covering $A$. You've got $l(I_0) \ge 1$ but only add it to the left-hand side of the inequality so your solution is in error.

Do you know that $m^*([0,1]) = 1$ and $m^*(rationals) = 0$? If so use subadditivity and monotonicity: $$m^*([0,1]) \le m^*(rationals) + m^*(irrationals) = m^*(irrationals) \le m^*([0,1])$$ so that $$m^*(irrationals) = m^*([0,1]) = 1.$$