A set $A$ is said to be almost contained in a set $B$ if $A\setminus B$ is finite. A sequence $(A_\alpha)_{\alpha<\lambda}$ of infinite subsets of $\mathbb N$ will be called a tower if for every $\alpha<\beta<\lambda$, $A_\beta$ is almost contained in $A_\alpha$. A tower is said to have a continuation if exists some infinite subset of $\mathbb N$ that is almost contained in every element of the tower. The tower number is defined to be the minimal cardinality of the set of towers that don't have a continuation.
My question is: why can't the tower number be $\aleph_0$? Or equivalently, why does every tower of countable length necessarily have a continuation?
This is a simple diagonalization argument. Suppose we have a countable tower. Taking a cofinal subsequence, we may assume it has length $\omega$ and write it as $(A_n)_{n<\omega}$. We may moreover assume that the $A_n$ are actually literally contained in each other instead of almost contained in each other, by replacing $A_n$ with $A_0\cap\dots\cap A_n$.
It's now easy to construct a set $B$ which is almost contained in each $A_n$: just pick one element from each $A_n$ to be in $B$. Picking these elements one by one, we can arrange that they are all distinct (since each $A_n$ is infinite), so that the resulting set $B$ is infinite. For each $n$, all but possibly the first $n$ elements we put in $B$ must be in $A_n$ (since they are in $A_m$ for some $m\geq n$), so $B$ is almost contained in $A_n$.