Suppose we have $\{I_{\alpha}, \alpha \in \mathbf{A}\ - arbitrary\ set\ of\ indexes \}$ - a family of non-intersecting intervals, $ I_{\alpha} \subset [0,1]$. How to prove that its cardinality is not more than countable?
UPD: I concluded, that, in fact, there is a more common statement with quite the same proof as proposed in my own answer. Concretely,
Let $\mathbb{H}$ be a separable metric space, B = $\{B_{\alpha}, \alpha \in \mathbf{A}\}$ - family of open balls in this space. Than the cardinality of B is not more than countable.
As I stated, the proof is quite the same, but instead of $\mathbb{Q}$ one must consider constructing a map from B to S, where S is a countable dense set in $\mathbb{H}$.
I just came to the proof. Each interval contains a rational number. So we can construct a map from the set of intervals to $\mathbb{Q}$. Intervals do not intersect, so this mapping is injective. So cardinality of the set of our intervals is the same as a cardinality of a subset of $\mathbb{Q}$.