$P$ is a prime ideal if $P$ satisfies the following : If $\bigcap\limits_{j=1}^{n}I_{j} \subseteq P$ for any ideals $I_1,I_2,..I_n$, then $I_j \subseteq P$ for some $j$, where $R$ is a commutative ring with unity:
I started off with let $xy \in P$, for $x$,$y \in R$. I need to create ideals such that whose intersection is precisely $xy$. I could think of $(x)$ and $(y)$ but I cann't guarantee that their intersection is precisely $xy$
Any hint on how to proceed
On
You always have $$I_1 \cdots I_n \subseteq \bigcap\limits_{i=1}^n I_i $$ Try to prove by induction that if $I_1 \cdots I_n \subseteq P$, one of the $I_i$ is contained in $P$ (some people take this as the definition of prime ideal). For example, here's the $n = 2$ case: let $IJ \subseteq P$, and suppose there is a $y \in J$ which is not in $P$. Then for any $x \in I$, $xy \in IJ \subseteq P$, so $x$ or $y$ is in $P$. But $y$ is not in $P$, so $x$ is in $P$. This shows that $I \subseteq P$ under the assumption that $J \not\subseteq P$.
When it's hard to get a grip on the logic of a problem, try the contrapositive. Suppose that $I_j\not\subset P$ for all $j$, so there are some $a_j \in I_j$ with $a_j \notin P$. And we are very nearly done.