Spectrum of a ring is irreducible if and only if nilradical is prime (Atiyah-Macdonald, Exercise 1.19)

I think the open sets definition of irreducibility is easier to work with. You should show these useful facts about $Spec(A)$: the sets $D(f) = \{\mathfrak{p} \in Spec(A) : f \notin \mathfrak{p}\}$ form a basis of the topology of the spectrum, and $D(f) \cap D(g) = D(fg)$. Then we suppose that $f \notin Nil(A)$ and $g \notin Nil(A)$. This means that $D(f)$ and $D(g)$ are then nonempty open sets, and so if $Spec(A)$ is irreducible, their intersection $D(fg)$ is nonempty.

We suppose that the nilradical is not prime and show that $\operatorname{Spec} A$ is reducible.

Let $\mathcal{N}$ be the nilradical of $A$. Suppose that $\mathcal{N}$ is not prime. Then there exist elements $a,b \in A$ such that $a,b \not\in \mathcal{N}$ but $ab \in \mathcal{N}$. Recall that $\operatorname{Spec} A = V(\mathcal{N})$ and $\mathcal{N} = \bigcap_{P \in \operatorname{Spec} A} P$. Next, define the sets $S_a = \left\{P \in \operatorname{Spec} A: \, a \in P \right\}$ and $S_b = \left\{P \in \operatorname{Spec} A: \, b \in P \right\}$. Notice that $S_a$ is non-empty (otherwise $b \in \mathcal{N}$) and proper subset of $\operatorname{Spec} A$ (otherwise $a \in \mathcal{N}$). Similarly for $S_b$. Writing $\mathcal{N} = \left(\bigcap_{P \in S_a} P \right) \cap \left(\bigcap_{P \in S_a^c} P\right)$, we have that $\operatorname{Spec}A = V(\mathcal{N}) = V\left(\bigcap_{P \in S_a} P \right) \cup V\left(\bigcap_{P \in S_a^c} P \right)$. It remains to show that $V\left(\bigcap_{P \in S_a} P \right)$ and $V\left(\bigcap_{P \in S_a^c} P \right)$ are proper subsets of $\operatorname{Spec}A$. Suppose that $\operatorname{Spec}A = V\left(\bigcap_{P \in S_a} P \right)$. Pick $P \in S_a^c$, then $P \in V\left(\bigcap_{P \in S_a} P \right)$ and so $a \in P$, contradiction. Next, suppose $\operatorname{Spec}A = V\left(\bigcap_{P \in S_a^c} P \right)$. Pick $Q \in S_b^c$. Then $Q \in V\left(\bigcap_{P \in S_a^c} P \right)$. But this is a contradiction, since $S_a^c \subset S_b$, and $b \in \bigcap_{P \in S_a^c} P$.

Let us denote the nilradical by $N$. $N= \cap P$. Let us assume N is not a prime ideal for X an irreducible space. So there must exist a and b in A, such that $a, b \not \in N$ but $ab\in N$. Then $ab \in P$ for all prime ideal P in A.

Now consider $X_{a} \cap X_{b}$, which is non-empty, by the property of irreducible space. Say some prime ideal, $P_{k} \in X_{a} \cap X_{b}$, $P_{k} \not \in (X_{a} \cap X_{b})^{c}= V(a) \cup V(b)$. That means $P_{k}$ is a prime ideal that do not contain a as well as b. That imply $ab \not \in P_{k}$. Which in turn shows $ab \not \in N$.