The union of family of open disjoint non empty subsets is dense. The statement I'm gonna write down here is a proposition I conjectured in order to prove a detail in Hartogs theorem; thus I won't write all the context, otherwise it would become the longest post ever in MSE! On the other side this statement is quite general, thus it shouldn't be a problem.
Consider the unitary open disk $\Delta$ in $\Bbb C$. $\Delta$ is a topological space, with topology induced by $\Bbb C$. We'll work in this topological space $\Delta$, thus, "open", "the closure of" etc are all related to this topological space.
Let $\Lambda$ be a family of index more than countable. Let $\{E_{\lambda}\}_{\lambda\in\Lambda}$ be a family of disjoint open non empty subsets of $\Delta$. Can we say that
$$
\operatorname{cl}_{\Delta}\left(\bigcup_{\lambda\in\Lambda}E_{\lambda}\right)=\Delta\;\;\;\;?
$$
I'd say yes, and it seems natural, but honestly I don't know neither where to start to prove it.
 A: I will show you that such a family of non-empty, disjoint open sets cannot exist.
To see this, let $\Lambda$ be an uncountable index set and suppose that $(E_{\lambda})_{\lambda\in\Lambda}$ is a family of disjoint, non-empty open subsets of $\Delta$. Also, let $$\mathbb Q_{\mathbb C}\equiv\{a+bi\,|\,\text{$a$ and $b$ are rational numbers}\}$$ be the set of numbers on the complex plane with “rational grid points” (that is, both the real and the imaginary parts are rational). It is not difficult to show that $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$ and also that $\mathbb Q_{\mathbb C}$ is countable.
For each $\lambda\in\Lambda$, pick $x_{\lambda}\in E_{\lambda}$ (you need the axiom of choice here). Since $E_{\lambda}$ is open and $\mathbb Q_{\mathbb C}$ is dense in $\mathbb C$, one can pick $q_{\lambda}\in\mathbb Q_{\mathbb C}$ such that $q_{\lambda}\in E_{\lambda}$. Because of disjointness, all the $(q_{\lambda})_{\lambda\in\Lambda}$ are distinct, so that the function $\lambda\mapsto q_{\lambda}$, mapping from $\Lambda$ to $\mathbb Q_{\mathbb C}$ is injective. This is implies that the set $\mathbb Q_{\mathbb C}$ is uncountable, which is impossible.
