From the book by Kenji Ueno, Algebraic Geometry 1. From Algebraic Varieties to Schemes:
"If an algebraic set $V(J)$ is reducible, it can be expressed as: $$(1.8)\quad V(J)= V(J_1)\cup V(J_2), \ V(J)\neq V(J_1),\ V(J)\neq V(J_2)$$
Hence, we have $V(J)\supsetneq V(J_j), \ j=1,2$, then we obtain from ex 1.11:
$$(1.9)\quad \sqrt{J}= I(V(J)) \subsetneq I(V(J_j))= \sqrt{J_j}$$
Therefore, there are polynomials $f_j\in \sqrt{J_j},\ j=1,2$, but $f_j \notin \sqrt{J}, \ j=1,2$. By $(1.9)$ we must have $f_1 f_2 \in \sqrt{J}$."
I don't understand why $f_1f_2 \in \sqrt{J}$, shouldn't it be in $\sqrt{J_1}\cdot \sqrt{J_2}$?
By $(1.8),$ we have $\sqrt{J} = \sqrt{J_1} \cap \sqrt{J_2}.$ Now we have $f_1 \cdot f_2 \in \sqrt{J_1} \cdot \sqrt{J_2} \subseteq \sqrt{J_1} \cap \sqrt{J_2} = \sqrt{J}.$
I think the last line is a misprint. it should read as "by $(1.8)$ we must have..." and I think that's where you had your problem.