For each $n\in\mathbb N_1$, set $E_n=|f|^{-1}[(\frac1n,\infty)]$ and $E_0=\varnothing$. Because $f$ is integrable, $\mu(E_n)<\infty$. Let $F=\bigcup_{n\in\mathbb N_1} E_n$ and $F_n=E_n\setminus E_{n-1}$. These sets form a disjoint partition of $F$ because $E_0\subseteq E_1\subseteq \ldots$, so from the countable additivity we have
$$\infty>\int_F|f|\,d\mu=\sum_{n=1}^\infty\int_{F_n}|f|\,d\mu,$$
which means for any $\varepsilon>0$ there's $n\in\mathbb N_1$ such that
$$\varepsilon>\sum_{k=n+1}^\infty\int_{F_k}|f|\,d\mu=\int_{F\setminus E_n}|f|\,d\mu=\int_{X\setminus E_n}|f|\,d\mu$$
because $f$ is $0$ on $X\setminus F$.
Let me first give accurate references to make citations less convoluted. The result you are citing is Exercise 4.6.44.(ii) from Royden & Fitzpatrick's Real Analysis, 4e (p. 96), while your lemmas (i), (ii) are Exercises 4.6.44.(i) (p. 95) and 3.2.18 (p. 63). Although to use these two results is suggested as a hint in the book, I believe there are less painful ways to do this approximation. I'll outline two of them.
1st way: First we prove this for $f:\mathbb{R}\to[0,\infty]$. Let $\varepsilon>0$. Like you started, first cut off the integral so that
$$\int_{\mathbb{R}-[-N,N]}f<\dfrac{\varepsilon}{2}$$
and then do an approximation by a simple function $s:[-N,N]\to[0,\infty[$ using 4.6.44.(i):
$$\int_{[-N,N]}|f-s|<\dfrac{\varepsilon}{4}.$$
Next, instead of weakening $s$ to be a measurable function, we consider it as is, and use Exercise 3.2.16 (p. 63) instead of 3.2.17:
Lemma (Exercise 3.2.16): Let $I\subseteq\mathbb{R}$ be a compact interval, $s:I\to\mathbb{R}$ be simple. Then $\forall\varepsilon>0,\exists$ step $\varphi:I\to\mathbb{R},\exists F\in\mathcal{M}:$
\begin{align}
&F\subseteq I, \forall x\in F: \varphi(x)=s(x),\\
&m(I-F)<\varepsilon.
\end{align}
Moreover we can bound $\varphi$ on $I$ by $M:=\Sigma_{i=1}^n|\alpha_i|$ provided that $s=\Sigma_{i=1}^n \alpha_i \chi_{E_i}$ is the canonical representation of $s$.
I added the last statement in the lemma, though it is immediate from the very construction of the required step function. Also observe that this answers your problem: we can have a bound on the integral over the compact interval that is dependent only on the values taken by the simple function $s$. (Consequently you were very close indeed.)
If we use the lemma to approximate $s$ by a step function with error $\dfrac{\varepsilon}{8M}$, $\int_{[-N,N]}|s-\varphi|<\dfrac{\varepsilon}{4}$. Noting that we can consider $\varphi$ to have the real line as its domain by $\varphi \chi_{[-N,N]}:\mathbb{R}\to\mathbb{R}$, we are done for nonnegative $f$.
For the general case it suffices to apply this argument to the positive and negative parts of $f$ (starting with half the error); everything adds up nicely.
2nd Way: Another way is to apply Exercise 4.3.21 (p. 85) directly to the positive and negative parts of $f$, after cutting off the integral outside a compact interval:
Lemma (Exercise 4.3.21): Let $I\subseteq\mathbb{R}$ be a compact interval, $f:I\to[0,\infty]$ be integrable. Then $\forall\varepsilon>0,\exists$ step $\varphi:I\to\mathbb{R}:m(\{\varphi\neq0\})<\infty,\int_E|f-\varphi|<\varepsilon$.
P.S. To be sure the first argument can be used to prove Exercise 4.3.21 so that I am really suggesting one solution only. That said I believe all these multilinks between exercises are mainly due to the step-by-step approach of the book.
Best Answer
Here, I will assume a slightly general version, namely when $f:\Bbb R\to \Bbb R\cup\{\pm \infty\}$ is Lebesgue integrable.
Consider $f_n:=\chi_{\{x\in \Bbb R:|x|>n\}}\cdot f$, then $f_n\to 0$ pointwise a.e. and $|f_n|\leq |f|$. So, Lebesgue Dominated convergence theorem gives $$\lim\int f_n =0.$$
Next, consider $g_n:= \chi_{\{x\in \Bbb R:|f(x)|>n\}}\cdot f$, then $g_n\to 0$ pointwise a.e. Note that we are using the fact here that $f$ is integrable implies $m\big(|f|^{-1}(\infty)\big)=0$, in case $f$ only takes real value then it is obvious. Also, $|g_n|\leq f$. So, Lebesgue Dominated convergence theorem gives $$\lim\int g_n =0.$$