Let $(X,d)$ be a metric space, and $K\subseteq X$.
Recall: A collection of open sets $\mathcal{G}\subseteq \tau_{X}$ in $X$ is called an open cover for $K$ if the following is true: $$K\subseteq \bigcup\mathcal{G}$$
Then I am interested in the following construction:
We are given a sequence of subsets of $X$, $\left(K_n\right)_{n\in\mathbb{N}}$, such that they form an ascending chain ordered by inclusion, i.e. for any $n$
$$ K_{n}\subseteq K_{n+1}$$
And also a sequence of open covers, $\left(\mathcal{G}_n\right)_{n\in\mathbb{N}}$, again an ascending chain ordered by inclusion, where each $\mathcal{G}_n$ is an open cover for $K_n$.
Then clearly
$$ \mathbf{K}:=\bigcup_{n\in\mathbb{N}}K_n\subseteq \bigcup\left(\bigcup_{n\in\mathbb{N}}\mathcal{G}_n\right) =: \mathbf{G}$$
Now if we further know that for each $n$, the cover $\mathcal{G}_n$ for $K_n$ is irreducible, in the sense that no proper subset of $\mathcal{G}_n$ is still a cover for $K_n$, then we can conclude that $\mathbf{G}$ is a finitely irreducible cover for $\mathbf{K}$, in the sense that no proper finite subset of $\mathbf{G}$ is a cover for $\mathbf{K}$. The proof for this is fairly simple:
If we suppose there is a proper, finite subset $\mathbf{G'}\subset \mathbf{G}$ which covers $\mathbf{K}$, then there is some $N\in\mathbb{N}$ such that $$\mathbf{G'} \subset \mathcal{G}_N $$ Contradicting the assumption that $\mathcal{G}_N$ is irreducible.$\hspace{1cm}\blacksquare$
However, this proof fails if we remove the "finite" requirement which makes me believe it might be necessary, but I have also not been able to find a counterexample. So my question is, can we conclude that $\mathbf{G}$ is not only finitely irreducible, but in fact irreducible? And as a follow up, if we define $\kappa$-irreducible for a cardinal $\kappa$ in the natural way, would we be able to conclude for example countably-irreducible if we started with sequences indexed by uncountable ordinals? Additionally if these ideas touch on things that have already been studied, references would be appreciated.
Counterexample. Let $X$ be the real line. For $n\in\mathbb N=\{1,2,3,\dots\}$ let $K_n=\bigcup_{k=1}^n(k-1,k)$ and let $\mathcal G_n=\{G_1,\dots,G_n\}$ where $G_n=(2^{-n},1)\cup(n-1,n)$. Then $\mathcal G_n$ is an irreducible open cover of $K_n$ but $\mathbf G=\bigcup_{n\in\mathbb N}\mathcal G_n$ is not an irreducible cover of $\mathbf K=\bigcup_{n\in\mathbb N}K_n$ since $\mathbf G\setminus\{G_1\}$ covers $\mathbf K$.