$\mathbb{R}^n$ Second Countable

Question

Lemma: Let $U$ be an open subset of $\mathbb{R}^n$ and $x \in U$ then there exists a ball with rational center $(\mathbb{Q}^n$ and rational radius containing $x,$ that is contained in $U.$

What I want to show: Every open subset of $\mathbb{R}^n$ is the union of some some collection of elements in the following set:

$\{$ Open balls with rational center and rational radius $\}$

Let $U$ be an arbitrary open set in $\mathbb{R}^n$. By the above lemma, for each $x \in U$ you can find a rational center $x'$ and rational radius $r'>0$ so that $x \in B(x',r')$ $\subset U$. Hence the following collection:

$B'= \{ B(a,r) : a \in \mathbb{Q}^n, r>0, B(a,r) \subset U \}$ gives us

$U= \bigcup_{B \in B'} B$

is the proof correct?

As this is part of the proof of showing that $\mathbb{R}^n$ is second countable, it follows that since every metric space is Hausdorff, $\mathbb{R}^n$ is Hausdorff, and trivially, it is locally euclidean of dimension $n$ (take the identity map, which is a homeomorphism) and so $\mathbb{R}^n$ is a $n$-dimensional topological manifold, am I correct?

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proofs that $\mathbb R^n$ is second countable and that $\mathbb R^n$ is an $n$- dimensional topological manifold are correct.