Show that every prime $p$ in $\textbf Z[i]$ divides some rational prime

I don't understand what what you wrote has to do with your question. But let $\omega$ be a prime element in $\mathbb{Z}[i]$. Then the ideal $\mathfrak p$ generated by $\omega$ is a prime ideal of $\mathbb{Z}[i]$. Hence $\mathfrak p \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. Thus either $\mathfrak p \cap \mathbb{Z} = 0$, or $\mathfrak p = p\mathbb{Z}$ for some rational prime number $p$.

The first case cannot happen, since $\omega$ is the root of an irreducible polynomial in $\mathbb{Q}[X]$ with coefficients in $\mathbb{Z}$, whose constant term is nonzero and lies in $\mathbb{Z} \cap \mathfrak p$.

You are forced into the second case, in which you have $p \in \mathfrak p = (\omega)$, which tells you that $\omega$ divides $p$ in $\mathbb{Z}[i]$.

We might not all be on the same page on what a rational prime is. Take a number like $7$, that's a prime number and a rational number, right? A number like $\frac 7 2$ is rational but it's not prime, not in $\mathbb Z$, anyway (it's not even in $\mathbb Z$ to begin with).

A number like $7 + 2 \sqrt 3$ is definitely not rational but it is a prime in $\mathbb Z[\sqrt 3]$. Now, what about a number like $7 + 2i$? It's definitely prime in $\mathbb Z[i]$, but is it rational? Some people say yes, others say no, that the mere fact that it is a complex number with a nonzero imaginary part resolutely excludes it from the rationals, regardless of the fact that the imaginary part multiplied by $-i$ (or by $i$, for that matter) is a rational real number.

But rather than get into that can of worms, let's just say that a number must be purely real before we can determine if it's rational or not. Therefore, the rational primes are $$\ldots, -13, -11, -7, -5, -3, -2, 2, 3, 5, 7, 11, 13, \ldots$$

However, some of these are actually composite in $\mathbb Z[i]$. Like $13$, for example: since $2^2 + 3^2 = 13$, then $(2 - 3i)(2 + 3i) = 13$. Then $2 + 3i$ is a divisor of $13$, which we can verify in Wolfram Alpha thus: 13/(2 + 3i) $$\frac{13}{2 + 3i} = 2 - 3i.$$ The same (being a divisor of $13$) is also true of $2 - 3i$.

Now take a number like $11$. There is no way to solve $x^2 + y^2 = 11$, at least not in integers of $\mathbb Z$. So the only divisors of $11$ are $-11, -1, -i, i, 1$ and... $11$ itself. A number trivially divides itself. So if $p \in \mathbb Z$ is a prime of the form $4k + 3$, that means it's a rational prime that divides itself.