If $E$ is measurable, then its complement $E^\complement$ is also measurable and we can find an open set $W$ so that $E^\complement \subset W$ and $m(W - E^\complement) < \varepsilon$. Let $F = W^\complement$ be a closed set. Since $E^\complement \subset W$, we have $F \subset E$. Moreover, $W-E^\complement = E-F$. Hence $m(E-F) < \varepsilon$.
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 $F$, $|f-h|=|f-\psi|<\varepsilon$.
Best Answer
Hint :
Let $\psi=\boldsymbol 1_A$ where $A\subset I$ is a measurable set. By Lusin theorem, if $\varepsilon>0$, there is an $F\subset I$ such that $m(I-F)<\varepsilon$ and $\psi|_F$ is continuous. Then $\psi|_F=0$ or $\psi|_F=1$, which is a step function.
I let you conclude if $\psi=\sum_{i=1}^n a_i\boldsymbol 1_{A_i}$.