I think $(3)$ is misleading. Since $E_k$ is the set of all sample points in exactly $k$ of the $A_i$ sets, $(3)$ makes no sense at all to me.
For $(3)$ I suggest we let each $E_k$ be a disjoint union:
$$E_k = \bigcup_{i=1}^{\binom{n}{k}} D_{k,i}$$
where each $D_{k,i}$ is the subset of $E_k$ where the $k$ $A_i$'s are a distinct $k$ of the sets $A_1,\ldots,A_n$.
Then, for $(3),\;$ we assume WLOG, for any given $k,i,\;$ that $D_{k,i}$ is contained in $A_1,\ldots,A_k$ (and in no others) and proceed as I think the author intends, with $D_{k,i}$ instead of $E_k$. Note that, by the additivity of probability:
$$P(E_k) = \sum_{i=1}^{\binom{n}{k}} P(D_{k,i}).$$
The idea of $(5)$ is that we have
\begin{eqnarray*}
P\left(\bigcup_{i=1}^{n}A_i\right) &=& \sum_{k=1}^{n} P(E_k) \qquad\qquad\text{by $(1)$} \\
&=& \sum_{k=1}^{n} \sum_{i=1}^{\binom{n}{k}} P(D_{k,i}) \qquad\text{by $(3)$}\qquad\qquad\qquad\text{(*)} \\
\end{eqnarray*}
By $(3),\;$ any $P(D_{k,i})$ appears in the expression $P_1-P_2+P_3-\cdots\pm P_k,\;$ exactly $\left(k-\binom{k}{2}+\binom{k}{3}-\cdots\pm\binom{k}{k}\right)$ times, and this equals $1$ by $(4)$.
So, from $(*),\;$ we can conclude that $P(\cup_{i=1}^{n}A_i) = P_1-P_2+P_3-\cdots\pm P_n,\;$ as required.
For each $\omega\in\Omega$ let $J(\omega)=\{j\in I:\omega\in A_j\}$, and note that $\omega\in S_{J(\omega)}$. In fact, $J(\omega)$ is the unique $J\subseteq I$ such that $\omega\in S_J$. To see this, let $J$ be any subset of $I$ different from $J(\omega)$, and suppose first that there is a $j\in J(\omega)\setminus J$. Then $\omega\in A_j$, so $\omega\notin\Omega\setminus A_j$; and by definition $S_J\subseteq\Omega\setminus A_j$, so $\omega\notin S_J$. Now suppose that there is a $j\in J\setminus J(\omega)$. Then $S_J\subseteq A_j$, but $\omega\in\Omega\setminus A_j$, so again $\omega\notin S_J$. Thus, $\omega\in J$ iff $J=J(\omega)$, and the sets $S_J$ are pairwise disjoint.
In fact each $S_J$ corresponds to one of the atomic regions in the Venn diagram. $S_\varnothing$, for instance, is the region outside all of the sets, and $S_I$ is the intersection of all of the sets. In a simple Venn diagram with $3$ sets, $A_1,A_2$, and $A_3$, $S_{\{1,3\}}$ is the set of points inside $A_1\cap A_3$ but outside $A_2$. Each of the atomic regions is uniquely identified by the collection of sets containing it: it is inside all of those and outside all of the rest.
Now suppose that $\omega\in\bigcap_{k\in K}A_k$. Then $K\subseteq J(\omega)$, and $\omega\in S_{J(\omega)}\subseteq\bigsqcup_{K\subseteq J\subseteq I}S_J$. Conversely, if $\omega\in\bigsqcup_{K\subseteq J\subseteq I}S_J$, then $K\subseteq J(\omega)$, and $\omega\in\bigcap_{k\in K}A_k$. Thus, $\bigcap_{k\in K}A_k=\bigsqcup_{K\subseteq J\subseteq I}S_J$.
Best Answer
It is enough to prove the inequalities of indicators $$1_{A} \leq \sum_{i}1_{A_i},$$ $$1_{A} \geq \sum_{i}1_{A_i} - \sum_{i < j}1_{A_i \cap A_j},$$ $$1_{A} \leq \sum_{i}1_{A_i} - \sum_{i < j}1_{A_i \cap A_j} + \sum_{i < j < k}1_{A_i \cap A_j \cap A_k},$$ etc. Because if these inequalities are shown, then evaluating and summing them over all $\omega \in A_1 \cup \dots \cup A_n$ finishes the proof.
So suppose $\omega \in A_1 \cup \dots \cup A_n$ is arbitray. So $\omega$ is in $k \geq 1$ of the sets. Then for odd $m$, we want to show that $$1 \leq k - \binom{k}{2} + \binom{k}{3} - \dots \pm \binom{k}{m},$$ and for even $m$ we want to show the reverse inequality. So we want to show that $$\sum_{i = 0}^{m}(-1)^i\binom{k}{i} \leq 0$$ for odd $m$, and the reverse inequality for even $m$. But this follows from the formula $$\sum_{i = 0}^{m}(-1)^i\binom{k}{i} = (-1)^m\binom{k - 1}{m}.$$