Assume that $R$ is a commutative ring.

Lemma. Let $A$ , $B$ , $C$ be ideals in $R$ such that $A\subseteq B\cup C$ . Then $A\subseteq B$ or $A\subseteq C$ .

Proof. Assume that this is not true. Then there are $x,y\in A$ such that $x\in B$ , $y\in C$ and $x\not\in C$ , $y\not\in B$ . Obviously $x+y\in A\subseteq B\cup C$ and without loss of generality we may assume that $x+y\in B$ . Then $y=(x+y)-x\in B$ . Contradiction. $\square$

Remark. This lemma is also true if we exchange ring with a group and ideals with subgroups (because we didn't use multiplication and commutativity of addition in proof).

Proposition. Let $I$ , $P_1,\ldots, P_n$ be ideals in $R$ such that each $P_i$ is prime. If $I\subseteq P_1\cup\cdots\cup P_n$ , then there exists $i\in\{1,\ldots,n\}$ such that $I\subseteq P_i$ .

Proof. We will use the induction on $n$ . For $n=2$ our lemma applies. Let $n>2$ . Assume that $I\not\subseteq P_1\cup\cdots\cup P_n$ . For $i\in\{1,\ldots, n\}$ define $$\overline{P_i}=P_1\cup\cdots\cup P_{i-1}\cup P_{i+1}\cup\cdots\cup P_n.$$ By our assumption (and induction hypothesis) $I\not\subseteq \overline{P_i}$ for any $i\in\{1,\ldots, n\}$ . Thus for any $i$ there is $x_i\in I$ such that $x_i\not\in\overline{P_i}$ .

Now for any $i\in\{1,\ldots, n\}$ define $\overline{x_i}=x_1\cdots x_{i-1} x_{i+1}\cdots x_n\in I$ . Then we have $$\overline{x_1}+\cdots +\overline{x_n}\in I$$ and thus there is $j\in\{1,\ldots, n\}$ such that $\overline{x_1}+\cdots +\overline{x_n}\in P_j$ . Since $\overline{x_i}\in P_j$ for any $i\neq j$ , then we have that $$\overline{x_j}\in P_j.$$ But $P_j$ is prime, so there is $k\neq j$ such that $x_k\in P_j\subseteq \overline{P_k}$ . Contradiction. $\square$

Counterexample. We will show, that if $P_i$ 's are not prime, then the thesis no longer hold, even when $n=3$ . Consider the ring of polynomials in two variables over a simple field of order $2$ , i.e. $\mathbb{Z}_{2}[X,Y]$ . Let $R=\mathbb{Z}_{2}[X,Y]/(X^2, XY, Y^2)$ . For $W(X,Y)\in \mathbb{Z}_{2}[X,Y]$ we shall write $\overline{W(X,Y)}=W(X,Y)+(X^2,XY,Y^2)\in R$ . Then it is easy to see, that $$R=\{ \overline{0}, \overline{1}, \overline{X}, \overline{Y}, \overline{X}+\overline{Y}, \overline{X}+\overline{1}, \overline{Y}+\overline{1}, \overline{X}+\overline{Y}+\overline{1}\}.$$ Let $$I=\{\overline{0}, \overline{X}, \overline{Y}, \overline{X}+\overline{Y}\};$$ $$A_1=\{\overline{0}, \overline{X}\};$$ $$A_2=\{\overline{0}, \overline{Y}\};$$ $$A_3=\{\overline{0}, \overline{X}+\overline{Y}\}.$$ It can be easily checked, that $I, A_1, A_2, A_3$ are all ideals and $I\subseteq A_1\cup A_2\cup A_3$ but obviously $I\not\subseteq A_i$ for any $i=1,2,3$ . $\square$