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.
This problem is badly written. One needs to specify how the three positive real numbers are chosen. Here are three interpretations, all of which I would consider reasonable, and which give very different answers to this question:
Interpretation 1 Choose $a$, $b$ and $c$ uniformly and independently in $[0,1]$. Then the probability that this condition holds is $1/2$.
Proof: We can think of the point $(a,b,c)$ as uniformly chosen in the cube $[0,1]^3$. The subset of $[0,1]^3$ where $a>b+c$ is a triangular pyramid with height $1$ and base of area $1/2$, so its volume is $1/6$; the same is true for $b>a+c$ and $c>a+b$. So the probability that, instead $a+b>c$, $a+c>b$ and $b+c>a$ is $1-1/6-1/6-1/6 = 1/2$.
Presumably, this is the interpretation the textbook intends. But I think that, at least equally reasonable, are the following:
Interpretation 2 Fix $a+b+c = N$. Sample $(a,b,c)$ uniformly at random in the triangle $\{ (a,b,c) : a,b,c \geq 0,\ a+b+c=N \}$. Then the region where $a+b>c$, $a+c>b$ and $b+c>a$ is a smaller triangles whose vertices are the midpoints of the edges of the original triangle, and thus has area $1/4$. So the probability is $1/4$.
You will get the same $1/4$ if you sample $(a,b,c)$ uniformly in the pyramidal region $\{ (a,b,c) : a,b,c \geq 0,\ a+b+c \leq N \}$, or if you sample $a$, $b$, $c$ independently at random for an exponential probability distribution.
Interpretation 3 Fix $a^2+b^2+c^2 = R^2$. Sample $(a,b,c)$ uniformly at random in the spherical triangle $\{ (a,b,c) : a,b,c \geq 0,\ a^2+b^2+c^2=R^2 \}$. This spherical triangle is $1/8$ of a sphere, so it has area $\tfrac{4 \pi}{8}R^2 = \tfrac{\pi}{2}R^2$. The region where $a+b>c$, $a+c>b$ and $b+c>a$ is a smaller spherical triangle all of whose angles are $\cos^{-1} (1/3)$, so it's area is $(3 \cos^{-1}(1/3)-\pi) R^2$ and the probability is $\tfrac{3 \cos^{-1}(1/3)-\pi}{\pi/2} \approx 0.35$.
You get this same $\tfrac{3 \cos^{-1}(1/3)-\pi}{\pi/2}$ if you sample from uniformly the region $\{ (a,b,c) : a,b,c \geq 0,\ a^2+b^2+c^2 \leq R^2 \}$, or if you sample $a$, $b$, $c$ independently at random from one-sided Gaussian distribution.
This could be a very nice textbook example to illustrate how answers to problems in probability depend on how they are formulated, and could lead to good class discussion of the best formulation, but I can't endorse your textbook singling out Interpretation 1 as the right one without discussion.
Best Answer
Letting $w_n$ be the number of permitted configurations we get the recursion
$$ \begin{align} w_n &= 1 + 3(n-1) + 3 \binom{n-1}{2} + w_{n-1}\\ \tag1 &=w_{n-1} + 1 + \frac{3}{2} n (n-1) \end{align} $$
Explanation: in the first line of $(1)$, each term in the RHS corresponds to the configurations that have $3,2,1,0$ values equal to $n$. The initial condition is $w_1=1$, or also $w_0=0$.
Noting that $p_n = w_n/n^3$, this concides with your recursion.
To solve this, one can postulate $w_n = a_1 n + a_2 n^2 + a_3 n^3$, replace on $(1)$ and solve for $a_i$. (This also can be attacked via generating functions, see eg). Or, noticing that $w_n - w_{n-1}$ is the discrete analog of the derivative, we can integrate the other side:
$$w_n - w_{n-1} = g_n \implies \sum_{k=1}^n g_k + w_0 = w_n $$
Then $$\begin{align} w_n &= \sum_{k=1}^n \left[1 + \frac{3}{2} k (k-1) \right] \\ &= \sum_{k=1}^n 1 + 3 \sum_{k=2}^n \binom{k}{2} \\ &= n + 3 \binom{n+1}{3} {\hskip 1cm} \\ &= \frac12 n^3+\frac12 n \end{align} $$
where we've used the Hockey-stick identity.
See also OIES A006003 where many alternative interpretations and results about this sequence are given.
The desired probability is then
$$p_n = \frac{w_n}{n^3}=\frac12 + \frac{1}{2n^2}$$