I found this problem in a book:
Let $(I_\lambda)_{\lambda \in \Lambda}$ be a not degenerate family of disjoint sets of $\Bbb{R}$. Prove that $card((I_\lambda)) \leq \aleph_0$ (That is, the family is at most denumerable
I have proved it by taking a rational number $q_\lambda$ such that $\forall \lambda \in \Lambda: q_\lambda \in I_\lambda \cap \Bbb{Q}$. Such $q_\lambda$ exists because $\Bbb{Q}$ is dense in every non degenerate interval of $\Bbb{R}$. Then I've built the function $f:(I_\lambda)_{\lambda \in \Lambda} \rightarrow \Bbb{Q}$ such that $I_\lambda \longmapsto q_\lambda$. The function is obviously injective because $\forall \lambda_1, \lambda_2 \in \Lambda, \lambda_1 \neq \lambda_2, I_{\lambda_1} \cap I_{\lambda_2} = \emptyset$ by hypothesis, hence showing that $card((I_\lambda)_{\lambda \in \Lambda}) \leq card(\Bbb{Q}) = \aleph_0$
But now I'm wondering whether the hypothesis are too restrictive: I can replace the disjoint part of the theorem with the following $$\forall \lambda_1, \lambda_2 \in \Lambda, \lambda_1 \neq \lambda_2, I_{\lambda_1} \cap I_{\lambda_2} \text{is non degenerate interval of $\Bbb{R}$}$$
EDIT2: neither of the weaker hypothesis below works. See accepted answer and comments
EDIT: As pointed out by JDMan4444 and Hagen von Eitzen this weaker condition does not hold but then I produced another condition weaker than the one given in the theorem but stronger than the one I've given:
$$\forall \lambda_1, \lambda_2 \in \Lambda, \lambda_1 \neq \lambda_2, I_{\lambda_1} \nsubseteq I_{\lambda_2} \; \text{and} \; I_{\lambda_1} \cap I_{\lambda_2} \text{is not degenerate interval of $\Bbb{R}$}$$
And the proof would be similar, but the $q_\lambda$ is taken from the interval that is unique to each set.
Actually, while I'm writing this question, I'm thinking that if we have a family of non degenerate intervals of $\Bbb{R}$ defined with the weaker hypothesis, we could define another family $$(J_\lambda)_{\lambda \in \Lambda} = \bigcap_{\lambda_1 \neq \lambda_2 \in \Lambda} (\Bbb{R} \smallsetminus (I_{\lambda_1} \cap I_{\lambda_2}))$$
of disjoint intervals of $\Bbb{R}$ which is denumerable as proven above... I don't know if this helps nor if it is correct
I don't think you can weaken the hypothesis as you have suggested.
Take $\Lambda=\mathbb{R}^{+}$, the non-negative reals, and define $I_{\lambda}=(-\infty,\lambda)$ for $\lambda\in\Lambda$. Pairwise intersections are non-degenerate, and there are uncountably many of them.
As I was typing this a comment was made with the same basic idea.
As per the comment below and edit above, we can define $I_\lambda=(-1/\lambda,1+\lambda^2)$ for $\lambda\in\mathbb{R}_{>0}$, the positive reals, showing that the issue is not that the intervals are nested.