union of infinitely many prime ideals
I fixed your notation so that there wasn't any equivocation. I changed $P_i$ to $m_i$, changed the ideal $I$ to $P$, and kept the set $I$ as $I$.
There is indeed a counterexample. Take $R=\mathbb C[x,y]$, the prime ideal $P=(x,y)$ is not contained in any of the ideals $m_{a,b}=(x-a,y-b)$ for $(a,b)$ not both $0$, since $P$ is also a maximal ideal. Yet every element of $P$ is contained in at least one of the $m_i$, because there is no polynomial function on $\mathbb A^2$ that vanishes on one point but not any other point. Thus $P \subset \bigcup_i m_i$.
On a related note, however, there are some important classes of rings $R$ where countable prime avoidance holds. That is, when an ideal $J$ is a subset of a countable union of prime ideals in such a ring, it has to be contained in a single element of the collection. In particular, $R$ satisfies countable prime avoidance if either:
$R$ contains an uncountable field, or
$R$ is a complete Noetherian local ring.
Case 1 was mentioned in an article by Hochster and Huneke in the Michigan Math. Journal in 2000. Case 2 was proved by Lindsay Burch in 1972. Both arguments are fairly straightforward.
I learned these things from the Hochster-Huneke article some years back.
As another counterexample, take any noetherian local ring $(R,\mathfrak{m})$ of dimension $>1$, such that $\mathfrak{m}\not\in\mathrm{Ass}(R)$. Then $\mathfrak{m}$ is subset of the union of all non maximal prime ideals, because any $x\in\mathfrak{m}$ lies in a prime ideal of height $\leq1$.