Why does at least two intervals overlap in an uncountable family of intervals?

Question

1. Prove that there are uncountably many intervals $(a,b)$ in $\mathbb{R}, a\neq b$.
2. Assume $X$ be an uncountable family of intervals. Show that there exists at least two intervals in this family that overlap.

First was not difficult. I used the arguments similar to Cantor's Diagonal Argument (used to show $\mathbb{R}$ is uncountable.)

My attempt for 2: Assume $X$ be an uncountable family of pairwise disjoint intervals, i.e. $(a_i,b_i) \cap (a_j,b_j) = \emptyset, \quad \forall i\neq j\in I$. We know there exists a rational number in each of these intervals. This implies there are uncountably many rational numbers. Contradiction, since $\mathbb{Q}$ is countable. Thus, $X$ must have at least two intervals that overlap. $\blacksquare$

Is there any problem with this reasoning?

Answer 1

Nope, no problems.

For (1), you certainly could use a diagonal argument directly to prove that there is no surjection from $\mathbb{N}$ onto the set of intervals, or as Git Gud points out you could instead use the existence of a surjection from the set of intervals to $\mathbb{R}$ and then appeal to the nonexistence of a surjection $\mathbb{N} \to \mathbb{R}$. It is common to use Cantor's theorem on the uncountability of the reals as a "black box" in this way.

Your proof for (2) is perfectly fine. You could also get the contradiction by showing that $X$ is countable after all, rather than by showing that $\mathbb{Q}$ is uncountable, but this choice is just a matter of taste.