Second Countability and Countability Problem:
Every collection of disjoint open subsets of a second countable topological space $X$ is countable.
My attempt:
Let $B_1,B_2,B_3....$ be an enumeration of $\mathbb{B}$, where $\mathbb{B}$ is the countable basis the topology on $X$ admits.
Let $M$ be an arbitrary collection of disjoint open sets of $X$.
If $M$ was finite then it is trivially countable.
Otherwise, since $U$ is a collection of open sets, each open set can be expressed as the union of some collection of elements in $\mathbb{B}$. So for every $U$ in $M$, $U$ is the union of some countable collection of element of $\mathbb{B}$. Hence set up a map $f: \mathbb{B} \rightarrow M$ to be such that $f(B_j)=U$, where $B_j$ is the element in $\mathbb{B}$ with the smallest index j in the union which expresses $U$. This is clearly surjective and so since $\mathbb{B}$ is countably infinite, so is $M$.
(1)Is this proof correct and complete? 
(2)What would be an alternative method?
Please answer the first one, if you are able to, answer the second.
 A: I find your proof a little unclear. The way you define $f$ doesn't quite work. 
Here's my try:
Since the elements of $M$ are disjoint,  there is an injection $i:M\to \Bbb B$ by choosing for each element $V$ of $M$ an element $U_V$ of $\Bbb B$ with $U_V\subset V$.  This can be done because $\Bbb B$ is a basis.   (We may need the axiom of choice here. )  Thus $\mid M\mid\le\mid\Bbb B\mid$ and $M$ is countable. 
A: *

*The strategy you're pursuing is not unworkable (see point 2 which does something similar from the other direction), but as it stands the proof is not "correct and complete". For example, the function $f$ you claim to be defining is not generally a function. Let $M$ consist of infinitely many disjoint subintervals of $(0,1)$ in $\mathbb{R}$ equipped with the usual topology, and consider the basis of open intervals with rational endpoints. What would your $f$ assign to the basic open $(5,6)$?

*For any open set $S$ of $X$ we can find some basic open $S_B \in \mathbb{B}$ such that $S_B \subseteq S$. So take any function $f: M \rightarrow \mathbb{B}$ that assigns to each open set in the collection $M$ a basic open contained within it. Since the open sets are pairwise disjoint, the corresponding basic opens are pairwise disjoint as well: this implies that $f$ is injective, and its codomain is countable, so we're done. If you wish, you can even fix the enumeration $B_1,B_2,\dots$ of $\mathbb{B}$ as above, and let $f(U)$ denote the least $B_j \subseteq U$.
A: First note that $X$ has a countable dense subset (i.e. $X$ is separable) $D=\{d_n: n \in \Bbb N\}$ by picking $d_n \in B_n$ for all $n$. The required property already holds for separable spaces (the property is called ccc, BTW)
Now if $\mathcal{U}$ is any disjoint family of non-empty open sets, every $U \in \mathcal{U}$ contains some unique $d(U)$ (by disjointness, different $U$ cannot contain the same one), if you want a definite choice, choose the $d_n$ with minimal index, and this defines a function $f: \mathcal{U} \to D$ which is injective, so $\mathcal{U}$ is (at most) countable.
If you want to work directly from your base and avoid the countable choice to get $D$, map $U \in \mathcal{U}$ to $B(U):=B_n$ with $n = \min \{k: B_k \subseteq U\}$ which is well-defined by well-orderedness of $\Bbb N$ and the fact that the set $\{k: B_k \subseteq U\}$ is non-empty, by $U$ being non-empty and $B_n, n \in \Bbb N$ being a base. Again $\mathcal{U} \ni U \to B(U) \in \Bbb B$ is 1-1.
