I was going through multiple proofs of Zorn's Lemma (Halmos', in Naive Set Theory, being amongst them), but I get stuck on the definition of towers, particularly the "union requirement" (requirement III in Halmos' proof and Proof Wiki).
Say $(X,\leq)$ is the poset under consideration and $\mathcal{X}$ is the set of all chains in $X$, partially ordered by inclusion. Let $\mathcal{N}$ be a subset of $\mathcal{X}$, and a candidate for being a tower. Now, the requirement I'm talking about is, if $\mathcal{K}\subseteq\mathcal{N}$ is a chain, then
$$\left(\bigcup_{K\in\mathcal{K}}K\right)\in\mathcal{N}.$$
What I don't understand is why this does not follow from the previous premises.
$\mathcal{N}$ is partially ordered under inclusion. Thus any chain $\mathcal{K}\subseteq\mathcal{N}$ must be totally ordered under inclusion. Doesn't that mean that there is a chain in $\mathcal{K}$ that already equals the union of all chains in $\mathcal{K}$ (so that said union is also in $\mathcal{N}$)?
For example, if $\mathcal{N} = \wp(\{1,2,3\})$ (the powerset of $\{1,2,3\}$), and say $\mathcal{K}=\{\{2\},\{2,3\},\{1,2,3\}\}$, then the union of all chains in $\mathcal{K}$ is $\{1,2,3\}\in\mathcal{K}$.
So, when would such union not be in $\mathcal{K}$? I searched for similar questions here and for other proofs online, but none seem to address this, which makes me think it's probably very obvious to everyone. Maybe I'm constantly misinterpreting something. Regardless, an explanatory example or just an explanation would be much appreciated.
Let $X=\mathcal{P}(\mathbb{N})$ and $\leqslant$ be set inclusion. Now, take$$\mathcal{K}=\bigl\{\{1\},\{1,2\},\{1,2,3\},\ldots\bigr\}$$and let $\mathcal N$ be the set of all finite subsets of $\mathbb N$. Then $\mathcal K$ is totaly ordered, right?! Besides, $\mathcal{K}\subset\mathcal N$. However, the union of all elements of $\mathcal K$ doesn't belong to $\mathcal N$.