Here is Theorem 1.5.8 in "Understanding Analysis" by S. Abbott, 2nd edition.

Theorem 1.5.8

(i) If $A_1, A_2, ..., A_n$ are each countable sets, then the union $A_1\cup A_2 \cup ... \cup A_n$ is countable.

(ii) If $A_n$ is a countable set for each $n \in \mathbb{N}$, then $\bigcup^\infty_{n=1}A_n$ is countable.

Exercise 1.5.3 (b) asks:

b) Explain why induction cannot be used to prove part (ii) of Theorem 1.5.8 from part (i).

On Slader I find this solution. I have pasted it below:

I don't quiet understand the reasoning above and why should it imply (does it?) that the infinite union would not be countable.

Suppose I split $\mathbb{N}$ into an infinite number of sets corresponding to the rows below:

$1 \;\;\;3 \;\;\;6 \;\;\;10\;\;...$

$2 \;\;\;5\;\;\; 9\;\;...$

$4\;\;\; 8\;\;...$

$7\;\;...$

The infinite union is clearly countable.

However, if I am to replicate the argument given in the solution, given $N\in \mathbb{N}$, we can always find infinitely many $x \in \bigcup_{n=1}^\infty A_n$ such that $x \not\in \bigcup_{n=1}^N A_n$.

What am I missing?