Countable basis of topology

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I know that the family $$\tilde{I}=\bigg\{(a,b)\;|\;-\infty<a<b<+\infty\bigg\}$$ is a basis of a single topology on $\mathbb{R}$, that is $\mathcal{T}(\mathbb{R})$ and the family

$$\overline{I}:=\{(a,b)\;|\;-\infty<a<b<+\infty\}\cup\{[-\infty,b)\;|\;-\infty<b<+\infty\}\cup\{(a,+\infty]\;|\;-\infty<a<+\infty\}$$ is a basis of a single topology on $\mathbb{\overline{R}=\mathbb{R}\cup{\{\pm\infty\}}}$ that is $\mathcal{T}(\mathbb{\overline{R}})$.

Question

How can I claim that these bases are countable?

If I consider the family $$\tilde{I_1}=\bigg\{(a,b)\;|\;a,b\in\mathbb{Q}\;,-\infty<a<b<+\infty\bigg\},$$

then $\tilde{I}\supseteq \tilde{I_1}$, does the vice versa also holds?

In the same way if I consider the family $$\overline{I_1}:=\{(a,b)\;|\;a,b\in\mathbb{Q}\;,-\infty<a<b<+\infty\}\cup\{[-\infty,b)\;|\;-\infty<b<+\infty\}\cup\{(a,+\infty]\;|\;-\infty<a<+\infty\}$$ then, $\overline{I}\supseteq \overline{I_1}$, does the vice versa also holds?

Thanks!