For your second question, we can use a generalization of your first proof. We note that $\mathbb{R}^n$ is a separable metric space, since $\mathbb{Q}^n$ is a countable dense subset. For metric spaces, this implies that $\mathbb{R}^n$ is second-countable.
This implies that there exists some countable collection $\mathcal{U} = \{U_i\}_{i=1}^\infty$ of open subsets of $\mathbb{R}^n$ such that any open subset of $\mathbb{R}^n$ can be written as a union of elements of some subfamily of $\mathcal{U}$. This is called a countable basis of $\mathbb{R}^n$ and each $U_i$ is called a basis element.
Then for each $x\in I$, since $x$ is isolated, then $\{x\}$ is open (a neighborhood of $x$ does not contain any other points, hence an open set containing $x$ does not contain other points, so $\{x\}$ must be open). Hence $\{x\}$ is a union of $U_i\in\mathcal{U}$, since all open sets are a union of the basis elements. But the only way this can be true is if $\{x\} \in \mathcal{U}$, since you can't have $\{x\}$ being a union of sets that all either contain other points than $x$ or are the empty set. The map $I\to\mathcal{U}:x\mapsto \{x\}$ is injective. Since $\mathcal{U}$ is countable, $I$ is countable.
This implies as a generalization that any set of isolated points in a second-countable space is countable.
