Every collection of disjoint non-empty open subsets of $\mathbb{R}$ is countable? This is my first question on the forum. I'm wondering if the following proof is valid.

Proof:
Let $\{A_\lambda\}_{\lambda \in L}$ be an arbitrary collection of disjoint non-empty open subsets of $\mathbb{R}$. Since every non-empty open subset of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals, we can take the union $A = \bigcup\limits_{\lambda \in L}A_\lambda$ and decompose it $A = \bigcup\limits_{n \in \mathbb{N}} I_n$ in disjoint open intervals which forms a countable collection. We can also decompose each $A_\lambda$ as $\bigcup\limits_{m \in \mathbb{N}}J_{\lambda,m}$. For $\lambda \neq \mu \in L$, $A_\lambda \cap A_\mu = \emptyset$ and this is a new representation of $A$:
$$A = \bigcup_{n \in \mathbb{N}} I_n = \bigcup_{\substack{\lambda \in L \\ m \in \mathbb{N}}} J_{\lambda,m}$$
No matter how complicated the union over $L$ is, the $J_{\lambda,m}$ are disjoint open intervals. Thus, by the uniqueness, the two collections are exactly the same. As the final argument, we produce an injection $\varphi:\{A_\lambda\} \mapsto \{J_{\lambda,m}\}$ picking for each $A_\lambda$ some $J_{\lambda,m}$.

I can't see any fault, but the result seems incredibly strong to me.
P.S.: stack exchange has some bug related to \bigcup and \bigcap symbols?
 A: To each of those open disjoint subsets you can associate one and only one rational number (just pick a rational number in the set). Thus you obtain an injection from your family of subsets into the set of rational numbers, which is countable. The conclusion follows that your family must indeed be countable.
A: It seems to me that:

"...every non-empty open subset of $\mathbb{R}$ can be written uniquely as a countable union of disjoint open intervals..."

is essentially what you are trying to prove (in fact, it's stronger than what you are trying to prove; it includes a uniqueness clause). So I would be very wary of using it. 
In addition, any argument that argues by saying "No matter how complicated the union..." is likely to be at least a little bit informal.
The result you are trying to prove can be established without invoking that rather strong result. 
Here's a hint: since the rational numbers are dense in $\mathbb{R}$, if $A$ is any open subset of $\mathbb{R}$, then $A\cap\mathbb{Q}\neq\emptyset$. Can you see why this observation yields the desired result?
A: Your argument seems valid to me, though I think that in assuming that every open subset of $\mathbb{R}$ has a unique representation as a countable disjoint union of open intervals, you're swatting a fly with a sledgehammer.
How about this?  In a disjoint collection $\{U_i\}_{i \in I}$ of nonempty open subsets, choose a rational number $x_{i}$ in each $U_i$.  then you get an injection $I \hookrightarrow \mathbb{Q}$, so $I$ is countable.  (Really this argument works in any separable topological space, i.e., whenever you have a countable dense subset.)
