If $C = (A_i)_{i\in I}$ is collection of disjoint sets, $A_i\subset \mathbb{R}^m$ open and not empty, then $C$ is countable If $C = (A_i)_{i\in I}$ is collection of disjoint sets, $A_i\subset \mathbb{R}^m$ open and not empty, then $C$ is countable
Attempt:
We prove that each set $A_i$ contains a point which has only rational coordinates. Let $a\in A_i$ for some $i\in I$, say $a= (a_1,\dots,a_m)$. Since $A_i$ is open, it is true that $B(a;r)\subset A_i$ for some $r>0.$ We note that $a_i\in \mathbb{R}$, and since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each  $1\leq i \leq m$ there exists a rational $q_i$ such that: $$a_i<q_i<a_i+\frac{r}{\sqrt m}$$
If $x_i = (q_i,\dots,q_m)\in \mathbb{R}^m$, then $x_i\in A_i$, since:
$$d(a,x_i) = \sqrt{(q_1-a_1)^2+\dots+(q_m-a_m)^2}<\sqrt{m\frac{r^2}{\sqrt{m}^2}}=r$$
and hence $B(a;r)\subset A_i$ implies $x_i\in A_i.$
We use the argument before and consider the family $(x_i)_{i\in I}$ of points whose coordinates are all rationals, each $x_i \in A_i$. Since the union is disjoint, we must have $x_i\neq x_j$ for every $i,j\in I$, $i\neq j$. 
Now, i'm having problems in defining the function that enumerates each set. I was thinking about to take a coordinate of each point that is not a coordinate for every other and then consider the "natural" enumaration of it.
Any help?
 A: We prove that a open set $A\subset \mathbb{R}^m$ contains a point which has only rational coordinates. Let $a\in A$, say $a= (a_1,\dots,a_m)$. Since $A$ is open, it is true that $B(a;r)\subset A$ for some $r>0.$ We note that each $a_i\in \mathbb{R}$, and since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each  $1\leq i \leq m$ there exists a rational $q_i$ such that: $$a_i<q_i<a_i+\frac{r}{\sqrt m}$$
If $q = (q_1,\dots,q_m)\in \mathbb{R}^m$, then $q\in B(a,r)$, since:
$$d(a,q) = \sqrt{(q_1-a_1)^2+\dots+(q_m-a_m)^2}<r,$$
and  $B(a;r)\subset A$ implies $q\in A.$
Now  we use the argument before for every open set $A_i$ and consider the family $(x_i)_{i\in I}$ of points whose coordinates are all rationals, each $x_i \in A_i$. Since the union is disjoint, we must have $x_i\neq x_j$ for every $i,j\in I$, $i\neq j$. 
Now define the following function: $f: C \rightarrow \mathbb{Q}^m$ by $f(A_i) = x_i$. Clearly $f$ is injective, because the union is disjoint. Well, since the set $\mathbb{Q}^m$ is countable, it follows that $C$ is also countable.
A: Let $f:\mathbb N\to \mathbb Q^n$ be a bijection.  For $A\in C=\{A_i:i\in I\}$ let $g(A)=\min \{j: f(j)\in A\}.$ Then $g$ is a bijection from $C$ to a subset of $\mathbb N,$ so $C$ is countable.
