In fact, the inclusion $IJ \subseteq I \cap J$ is always true. Given ideals $I$ and $J$, an element of $IJ$ is of the form $x = a_1 b_1 + \cdots + a_t b_t$ for some $t \in \mathbb{Z}_{\geq 0}$ and $a_k \in I$ and $b_k \in J$. Since ideals are closed under arbitrary multiplication, then each term of the above sum is in both $I$ and $J$, and since ideals are closed under addition, then the sum is in $I$ and $J$, so $x \in I \cap J$.
Now, let's consider the reverse inclusion. For ease of notation, let's just consider two primary ideals $Q_1$ and $Q_2$. By part (b) we have $Q_1 = (p_1^m)$ and $Q_2 = (p_2^n)$ for some primes $p_1, p_2$ and, as per my comment, assume $p_1$ and $p_2$ are non-associate. Given $x \in Q_1 \cap Q_2$, then $x = p_1^m a$ and $x = p_2^n b$ for some $a,b \in R$. Then $p_1 \mid p_2^nb$, so $p_1 \mid p_2^n$ or $p_1 \mid b$. If $p_1 \mid p_2^n$, then by repeated using the fact that $p_1$ is prime we find that $p_1 \mid p_2$. Then $p_2 = p_1 c$ for some $c \in R$. But since primes are irreducible, then $c$ must be a unit, which contradicts that $p_1$ and $p_2$ are non-associate. Thus $p_1 \mid b$, so $b = p_1 b_1$ for some $b_1 \in R$.
Now we have
$$
p_1^m a = x = p_2^n b = p_2^n p_1 b_1 \, .
$$
Since $R$ is a domain, then we cancel the factor of $p_1$, which yields $p_1^{m-1} a = p_2^n b_1$. Repeating the same argument as above (or by induction, if you want to be rigorous), then $p_1 \mid b_1$ so $p_1^2 \mid b$, and so on and so forth until we get $p_1^m \mid b$. Then $b = p_1^m b_m$ for some $b_m \in R$, so
$$
x = p_2^n b = p_2^n p_1^m b_m \in (p_1^m)(p_2^n) = Q_1 Q_2 \, .
$$
We can use the above in the induction step to prove this result for any finite number of primary ideals. I'll leave the details to you.
Suppose we know the result for $k$ primary ideals. Then \begin{align*} Q_1 \cap \cdots \cap Q_k \cap Q_{k+1} &= (Q_1 \cap \cdots \cap Q_k) \cap Q_{k+1} = (Q_1 \cdots Q_k) \cap Q_{k+1} \, . \end{align*}
Given $x \in \bigcap_{i=1}^{k+1} Q_i = (Q_1 \cdots Q_k) \cap Q_{k+1}$, then $$x = p_1^{n_1} \cdots p_k^{n_k} a \qquad \text{and } \qquad x = p_{k+1}^{n_{k+1}} b $$ for some $a,b \in R$. You can use the same proof as above to show that $p_{k+1}^{n_{k+1}} \mid a$ which will finish the induction.
If we take an isomorphism, as stated in a comment, we get a polynomial ring over a field which is a UFD and the primes are irreducible polys over $\mathbb{Q}[\omega]$. There are plenty of them, because the field is not algebraically closed.
Best Answer
Yes, your proof is fine. Here is another way to view it. In a UFD every prime ideal $\rm\:P\:$ may be generated by primes. Indeed, $\rm\:0\ne p_1\cdots p_n\in P\:\Rightarrow\:$ some $\rm\: p_i\in P,\:$ so we may replace each generator by one of its prime factors. Hence if prime $\rm\:(p)\supseteq Q = (p_1,p_2,\ldots)\ne 0\ne p_i\:$ then $\rm\:(p)\supseteq (p_i)\:$ $\Rightarrow$ $\rm\:(p) = (p_i)\:$ $\Rightarrow$ $\rm\: Q = (p).\,$ QED $\, $ A converse is a famous theorem of Kaplansky.
Theorem $\ $ TFAE for an integral domain D
$\rm(1)\ \ \:D\:$ is a UFD (Unique Factorization Domain)
$\rm(2)\ \ $ In $\rm\:D\:$ every prime ideal is generated by primes.
$\rm(3)\ \ $ In $\rm\:D\:$ every prime ideal $\ne 0$ contains a prime $\ne 0.$
Proof $\ (1 \Rightarrow 2)\ $ Proved above. $\rm\ (2\Rightarrow 3)\ $ Clear.
$(3 \Rightarrow 1)\ $ The set $\rm\:S\subseteq D\:$ of products of units and nonzero primes forms a saturated monoid, i.e. $\rm\:S\:$ is closed under products (clear) and under divisors, since the only nonunit divisors of a prime product are subproducts (up to associates), due to uniqueness of factorization of prime products. Since $\rm\:S\:$ is a saturated monoid, its complement $\rm\:\bar S\:$ is a union of prime ideals. So $\rm\:\bar S = \{0\}\:$ (else it contains some prime ideal $\rm\:P\ne 0\:$ which contains a prime $\rm\:0\ne p\in P\subseteq \bar S,\:$ contra $\rm\:p\in S).\:$ Hence every $\rm\:0\ne d\in D\:$ lies in $\rm S,\:$ i.e. $\rm\:d\:$ is a unit or prime product. Thus $\rm\:D\:$ is a UFD. $\ $ QED
Remark $\ $ The essence of the proof is clearer when one learns localization. Then ones sees from general principles that prime ideals in $\rm\bar S\:$ correspond to maximal ideals of the localization $\rm\:S^{-1} D.\:$
In fact one can view this as a special case of how UFDs behave under localization. Generally the localization of a UFD remains a UFD. Indeed, such localizations are characterized by the sets of primes that survive (don't become units) in the localizations.
The converse is also true for atomic domains, i.e. domains where nonzero nonunits factor into atoms (irreducibles). Namely, if $\rm\:D\:$ is an atomic domain and $\rm\:S\:$ is a saturated submonoid of $\rm\:D^*$ generated by primes, then $\rm\: D_S$ UFD $\rm\:\Rightarrow\:D$ UFD $\:\!$ (popularized by Nagata). This yields a slick proof of $\rm\:D$ UFD $\rm\Rightarrow D[x]$ UFD, viz. $\rm\:S = D^*\:$ is generated by primes, so localizing yields the UFD $\rm\:F[x],\:$ $\rm\:F =\:$ fraction field of $\rm\:D.\:$ Therefore $\rm\:D[x]\:$ is a UFD, by Nagata. This yields a more conceptual, more structural view of the essence of the matter (vs. traditional argument by Gauss' Lemma).