Stuck on existence proofs involving measurability and simple functions

What you are missing in 1 is that you want to use your cover related to $E$ and not $I$.

Since $E$ is measurable and $E\subset I $, there exists an open cover $O=\bigcup_1^\infty I_k$ of $E$ such that $$m(O\setminus E)\leq\sum_km (I_k)-m (E)<\varepsilon/2.$$ As the sum of the series is finite, we may choose $n $ with $\sum_{k>n}m (I_k)<\varepsilon/2$. Put $O'=\bigcup_1^nI_k $. We have $$m (E\cap O')=m (E)-m (E\cap (O\setminus O')>m (E)-\varepsilon /2. $$

Let $h=\sum_1^n\chi_{J_k}=\chi_{O'}$, where $J_k=I_k\setminus \cup_1^{k-1}I_j $; each $J_k $ is a finite union of disjoint intervals, so $h $ is a step function. Let $F=(E\cap O')\cup (I\setminus O)$. Then $$ m(I\setminus F)=m(O\setminus (E\cap O'))\leq m (O)-m (E\cap O')<m (O)-m (E)+\varepsilon/2<\varepsilon. $$ And $h$ is $1$ on $E\cap O'$ and $0$ on $I\setminus O$, so $h=\chi_E$ on $F$.

For 2, remember that $\psi$ is simple, not step. So $\psi=\sum_1^na_k\chi_{E_k}$ with $E_k$ measurable. Now apply 1 to each $E_k$, to obtain $F_1,\ldots,F_n$ measurable and $h_1,\ldots,h_n$ step with $h_k=\chi_{E_k}$ on $F_k$, and $m(I\setminus F_k)<\varepsilon/n$. Let $h=\sum a_k h_k$ (sum of step is step). Let $F=\bigcap F_k$. Then $h=\psi$ on $F$, and $$ m(I\setminus F)\leq\sum m(I\setminus F_k)<\varepsilon. $$

For 3, fix $\varepsilon>0$. Since $f$ is bounded and measurable, there exists $\psi$ simple with $|f-\psi|<\varepsilon$. By 2, there exists $h$ step and a measurable set $F$ with $h=\psi$ on $F$ and $m(I\setminus F)<\varepsilon$. On $F$, $|f-h|=|f-\psi|<\varepsilon$.