Proposition 1.11 in Atiyah-Macdonald's "Introduction to commutative algebra" states the following: 

"Given an ideal $I$ in a ring $A$ and $p_1, \dots p_n$ prime ideals, then $I \subset \cup_i p_i$ implies that $I\subset p_j$ for some $j\in \{1, \dots, n\}.$ "

This is clearly equivalent to saying that "if $I$ is not a subset of $p_i$ for any $i,$ then $I$ is not a subset of $\cup_i p_i.$" 

But they then claim that this last statement is logically equivalent to the following: 

"For each $i,$ there exists $x_i \in I$ such that $x_i \notin p_k$ for all $k\neq i.$" 

Can anyone help me to see why this statement is logically a consequence of the first?  I know this is elementary, but I'm truly stuck and I'm hoping that someone can phrase things in a way that makes them more transparent. 


Question: Two different analytic curves cannot intersect in infinitely many points
Answer: Wrong: $t\mapsto \binom{t}{\sin(t)}$ and $t\mapsto\binom{t}{\cos(t)}$.

Edit: Grzegorz pointed out that $t$ is in a compact interval $[\alpha,\beta]$.
Under this assumption the statement is correct.

Proof: Let $f,g:[\alpha,\beta]\to \mathbb R^n$ be the two real analytically parameterized curves which intersect in $f(t_i)=g(t_i), i=1,2,\dots$ different points. Then the $t_i$ have accumulation points in $[\alpha,\beta]$, thus $f$ and $g$ coincide on $[\alpha,\beta]$.

In the $C^\infty$ case the statement is wrong: Let $[\alpha,\beta]=[0,1]$ and consider
$$f(t) = \binom{t}{e^{-1/t^2}\sin(1/t)},\qquad g(t)=\binom{t}{0}.$$

2nd edit: Noam pointed out that my proof above was not conclusive. 
So let me try again: Suppose that $f(t_i)=g(u_i)$ for sequences of distinct points.
The $t_i$ have accumulation an point $t$ and the $u_i$ have an accumulation point $u$.
By continuity, $x=f(t)=g(u)$. Now we change both parameterizations as follows:
Choose a line $r\mapsto x + r.v$ for a vector $v$ such that both curves are transversal to the hyperplane $v^\bot$. Since both curves are immersions (if I understood the question right),
such $v$ exists. Now consider the orthogonal projection of both curves onto this line. In a neighborhood of $r=0$ these are real analytic diffeomorphisms, so their inverses gives us new parameterizations of both curves in the parameter $r$. There are still infinitely many intersection points of the two curves near $x$, but these intersections happen now at the same 
parameter values $r_i$ with $r_i\to 0$ without loss. Now my proof from above applies. 

I hope that I did not overlook something else this time. Many thanks for pointing out my mistakes. 

3rd edit: I overlooked again something (Thanks SJR and Ramiro). So what I proved is:
The intersection of the two curves contains an open interval in each curve. Note that each curve, as an immersion, is locally a real analytic submanifold.  

4th edit: Grzegorz, to the new question you posed in the comment, the following comes to my mind:
By a theorem of Whitney, any closed set in $\mathbb R^n$ is the zero locus of a smooth function. So take a set in $\mathbb R$ like the Cantor set which in uncountable, and a smooth function on $\mathbb R$ vanishing exactly on this set. The graph of this function and the $x$-axis are two $C^\infty$-curves which intersect in this set.

What do mean by: $K_1\cap K_2$ is not $C^\infty$?

5th edit: Grzegorz, instead of excluding Ramiro's example in your edit of the question,
you could reformulate as follows:

Consider two  immersed curves which are parameterized real analytically on compact intervals.
If they have an infinite number of different intersection points, then their union is again a real analytic immersed curve.