Simple set-theoretic property to extract a subsequence in infinitely many steps Well, I know that I'm going to have some vote-down, because this should be a very simple property but the real story is that... I am not able to prove it!
For any fixed $n\in\mathbb N$, I have a finite partition of the natural numbers
$$
\mathbb N=A_1^n\cup...\cup A_{k(n)}^n
$$
(every set of the partition is infinite). Suppose that the sequence $k(n)$ is bounded. I want to make a choice of a set $A_{i_n}^n$ for each partition such that $A=\bigcap_{n=1}^\infty A_{i_n}^n$ is infinite.
Intuitively this should be possible, and I have a sketch of the proof making use of the upper density on natural numbers, but I don't think it is completely right and anyway it sounds a bit too complicated.
Motivation: I am basically extracting infinite subsequences from a sequence and I want to be sure to find, at the end of the story, a subsequence.
 A: As requested, here's the comment posted as an answer:
What if the $n$th partition splits the naturals into ten pieces based on the $n$th digit of the decimal expansion?
(Of course ten is chosen for familiarity and not for optimality.)

Apologies for such an extensive edit, but I couldn't visualize this either, and this doesn't fit into a comment. Using binary (and indexing from 0) instead of base 10 gives:
  $$A_0^0 = evens, A_1^0 = odds$$
  $$A_0^1 = (0\bmod 4)\cup(1\bmod 4), A_2^1 = (2\bmod 4)\cup(3\bmod 4)$$
  $$A_0^n = \{ n \in {\mathbb N} \colon 0\leq n\bmod 2^{n+1} <2^n\},
    A_1^n = \{ n \in {\mathbb N} \colon 2^n\leq n\bmod 2^{n+1} <2^{n+1}\},$$ 
But now $$\bigcap_{i=0}^n A_{i_n}^n$$ is a single congruence class modulo $2^{n+1}$ for any choice of $i_n$, and the second positive element of a congruence class must be at least as large as the modulus, and the modulus goes to infinity, so the infinite intersection cannot have two elements.
A: I think Clinton Conley intended for the $n$th digit to be read from the right (one's digit is the first, ten's digit is the second, etc.).  
