I'm a high school student self-studying analysis, and I'm reading Apostol's book. I wrote a proof that shows that $\mathbb{R}^n$ has a countable base, but am unsure if it is correct. I just wanted some feedback on it. Thanks!
Prove that $\mathbb{R}^n$ has a countable base.
Proof. We construct an open cover $\{V_{\alpha}\}$ for $\mathbb{R}^n$ by considering the union of all neighborhoods with rational centers and rational radii, namely $$\{V_{\alpha}\} = \{ N_q(n):q, n \in \mathbb{Q} \}$$ Pick an arbitrary $x \in \mathbb{R}^n$, and take an arbitrary open set $S$ such that $x \in S.$ Since $S$ is open, there must be a neighborhood $N_{\epsilon}(x)$ with radius $\epsilon$ about $x$ such that $N_{\epsilon}(x) \subset S$.
Take point $j \in \mathbb{Q}^n$ such that $d(j,x) < \frac{\epsilon}{100}$, which exists since $\mathbb{Q}^n$ is dense in $\mathbb{R}^n.$
Now, take $k \in \mathbb{Q}$ so that $ \frac{\epsilon}{100} < k < \frac{\epsilon}{10}$. Therefore, $N_k(j) \in \{V_{\alpha}\}$. Clearly, $$x \in N_k(j) \subset N_{\epsilon}(x) \subset S.$$ $\{V_{\alpha}\}$ is thus a base.
$\{V_{\alpha}\}$ is countable, since it is a union of a countable collection of countable sets.
Thus, $\{V_{\alpha}\}$ is a countable base.
