A nested cover of a set that eventually has infinite intersection with every infinite subset Prove the following
Let $\{A_n\}_{n \in \mathbb{N}}$ be a sequence of sets with $$A_1 \subset A_2 \subset A_3 \dots $$ and $B \subset \bigcup_{n = 1}^{\infty} A_n$. If for every infinite subset $E$ of $B$, there exists $A_i$ such that $A_i \cap E$ is infinite, then there exists $k_0 \in \mathbb{N}$ such that $B \subset A_{k_{0}}$
I became stuck on this exercise in my book after attempting it for a bit on paper. Could anyone help with this problem? I have been stuck on this for awhile. 
 A: If $B$ is finite, then you can check directly from $B\subset\bigcup_n A_n$ that the claim is true. So we can assume that $B$ is infinite.
Suppose the claim is false, then for each $k\geq 1$, let $a_k$ be in $B$ but not in $A_k$ and let $E=\bigcup_k\{a_k\}$. Claim that such $\{a_k\}$ can be chosen to be distinct so that $E$ is infinite:


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*The base case with just $a_1$ is trivial.

*Suppose that $a_1,\ldots,a_k$ are distinct. Then there has to be $a_{k+1}$ distinct from $a_1,\ldots,a_k$ such that $a_{k+1}\in B$ but $a_{k+1}\not\in A_{k+1}$. Indeed, otherwise we have $B\backslash\{a_1,\ldots,a_k\}\subset A_{k+1}$ and so $B\subset\bigcup_n A_n$ implies $B\subset A_N$ for some sufficiently large $N>k+1$ satisfying $\{a_1,\ldots,a_k\}\subset A_N$.


Because $E$ is infinite, there exists $A_i$ such that $A_i\bigcap E$ is infinite. This means that $A_i$ contains $a_j$ for $j>i$. But this implies that $a_j\in A_i\subset A_j$, a contradiction to the construction that $a_j\not\in A_j$.
A: Suppose that for all $i$, $B$ is not a subset of $A_i$, and build an infinite sequence that will form a set $E$ to which you can apply the hypothesis. 
