Showing that every collection of disjoint open subsets is countable Let X be a second countable topological space. Show that every collection of disjoint open subsets of X is countable. 
Proof: Let $X$ be a second countable space. Choose ${\{B_{i}\}{_i\in \mathbb{N}} }$ to be our countable basis for the topology on $X$. Let $C$ be any collection of disjoint open subsets of $X$. Our goal is to show that $C$ is countable. More formally, let $C=\{U_{i} \in \{B_{i}\}\subseteq X  | \: i \in \mathbb{N}, \: U_{i} \bigcap U_{j} = \emptyset \}$. Now $C$ is countable since for every $U_{i} \in \{B_{i}\} $ there is a correspondence between $i \in \{1, ..., n\}$ for $n \in \mathbb{N}$. Thus it follows that $C$ is countable. 
Is the construction of the set $C$ precise or is their a better way.
 A: In your construction of $C$, you're assuming it's countable (its elements are indexed by a countable set).  What you should do is start with an arbitary collection $C=\{U_\alpha\}$ of pairwise disjoint open subsets of $X$, then show that $C$ is countable.
Hint: Each $U_\alpha$ contains some $B_\alpha$, and since $U_\alpha\displaystyle\cap U_\beta=\varnothing$ if $\alpha\neq\beta$, we know that $B_\alpha\displaystyle\cap B_\beta=\varnothing$ if $\alpha\neq\beta$.
A: Suppose that $\{B_n: n \in \mathbb{N}\}$ is a countable base for $X$.
Then let $\mathcal{U}$ be any family of pairwise disjoint non-empty open sets.
For every $U \in \mathcal{U}$ find $x \in U$ and as we have a base, there is some $n(U) \in \mathbb{N}$ such that $x \in B_{n(U)} \subseteq U$.
This defines a function: $f: \mathcal{U} \to \mathbb{N}$ defined by $f(U) = n(U)$. If $U \neq V$ then $n(U) \neq n(V)$ otherwise $m = n(U) = n(V)$ and then $\neq = B_m \subseteq U_{n(U)} \cap U_{n(V)}$, a contradiction with the disjointness of $\mathcal{U}$. So $f$ is 1-1 and so $\mathcal{U}$ is at most countable.
