Trouble with proof that $L^p(\mathbb R ^n)$ is separable

I'm having trouble understanding the following proof from Linear Functional Analysis - An Application-Oriented Introduction (4.18, Proof for (4)):

Let $\epsilon>0, 0<p<\infty$ and $f\in L^p(\mathbb R ^n)$. Let further$$Q_\epsilon(z) := \{x \in \mathbb{R}^n ; z_i \leq x_i \leq z_i + \epsilon \ \text{for} \ i = 1, \ldots, n \} \ \text{for} \ z \in \epsilon \mathbb{Z}^n$$

We define the piecewise constant interpolation $f_\epsilon$ as:

$$f_\epsilon(x) := \sum_{z \in \epsilon \mathbb{Z}^n} \chi_{Q_\epsilon(z)}(x) \alpha_{\epsilon, z}, \ \ \alpha_{\epsilon, z} := \frac{1}{\epsilon^n} \int_{Q_\epsilon(z)} f(y) \, dy$$

This interpolant approximates $f$, since:$$\begin{align}\int_{\mathbb{R}^n} |f_\epsilon(x) - f(x)|^p \, dx & = \sum_z \int_{Q_\epsilon(z)} |\alpha_{\epsilon, z} - f(x)|^p \, dx\\& = \sum_z \int_{Q_\epsilon(z)} \left| \frac{1}{\epsilon^n} \int_{Q_\epsilon(z)} (f(y) - f(x)) \, dy \right|^p \, dx \label{1}\tag{1}\\& = \sum_z \frac{1}{\epsilon^n} \int_{Q_\epsilon(z)} \int_{Q_\epsilon(z)} |f(y) - f(x)|^p \, dy \, dx\label{2}\tag{2}\\& \leq \sup_{h : |h| \leq \epsilon} \int_{\mathbb{R}^n} |f(y + h) - f(y)|^p \, dy = \sup_{h : |h| \leq \epsilon} \| f(\cdot + h) - f \|_{L^p}^p.\label{3}\tag{3}\end{align}$$

I'm honestly totally lost: In \eqref{1} for example, I don't see how we can factor out $\frac{1}{\epsilon^n}$ like that. I tried doing it step by step, but I get:$$\begin{align}\sum_{z}\int_{Q_\epsilon(z)}\left|\alpha_{\epsilon,z}-f(x)\right|^p\, dx& =\sum_{z}\int_{Q_\epsilon(z)}\Bigg|\bigg(\frac{1}{\epsilon^n}\underbrace{\int_{Q_\epsilon(z)} f(y)\, d y}_{=:u}\bigg)-f(x)\Bigg|^p\, dx\\& =\sum_{z}\int_{Q_\epsilon(z)}\left|\frac{1}{\epsilon^n}\left(u-\epsilon^nf(x)\right)\right|^p\, dx\\& =\frac{1}{\epsilon^n}\sum_{z}\int_{Q_\epsilon(z)}\left|u-\epsilon^nf(x)\right|^p\, dx\\& =\frac{1}{\epsilon^n}\sum_{z}\int_{Q_\epsilon(z)}\left|\int_{Q_\epsilon(z)}{f(y)-\epsilon^nf(x)}\, d y\right|^p\, dx\end{align}$$But for $\epsilon\neq 1$ this seems distinct from \eqref{2}.

For \eqref{2} I do not understand why we have an equality, and why it holds for all $p$. I could get an inequality $\le$ since for any function $g$ we have $\left|\int_{\Omega} gd\mu\right|\le\int_{\Omega}\left|g\right|d\mu$ and we maybe can fit Jensen's inequality, but that'd only give an inequality?