Euclidean Space is Second Countable
# Statement
The Euclidean Space $\mathbb{R}^{n}$ (for $n \in \mathbb{N}$) is a Second Countable Topological Space.
# Proof
Consider the collection $$\mathcal{B} := {B_{\delta}(x) : \delta \in \mathbb{Q}, x \in \mathbb{Q}^{n}}$$ It is Countable (A Union of Countable Sets is Countable, Rationals are Countable, and Finite Products of Countable Sets dare Countable). It is a collection of Open sets by definition of Metric Topology. Let $U \subset \mathbb{R}^{n}$ be Open. Consider $V := \bigcup\limits_{B \in \mathcal{B}, B \subset U} B$. By construction, $V \subset U$. We will show $V \supset U$. Let $x \in U$. By definition of Metric Topology, there exists a $\delta \in \mathbb{R}{>0}$ so that $B{\delta}(x) \subset U$. Because the Rationals are Dense in the Reals, there exists a $\delta’ \in \mathbb{Q} \cap (0, \delta)$ (i.e. $0 < \delta’ \leq \delta$). Again, because the Rationals are Dense in the Reals, there exists an $x’ \in \mathbb{Q}^{n} \cap B_{\delta’}(x)$. Therefore, $x \in B_{\delta’}(x’) \subset V$. Since $x \in U$ was arbitrary, we have $V \supset U$.
Therefore $\mathcal{B}$ is a Topological Basis for $\mathbb{R}^{n}$, so $\mathbb{R}^{n}$ is Second Countable. $\blacksquare$