Prove that $\mathbb{R}^k$ is separable
Solution 1:
That is pretty much it. We can write $\mathbb{Q}^k=\{(x_1,...,x_k)\}$ and this is countable because the cartesian product of countable sets is countable ($\mathbb{Q}$ is countable). When constructing your arbitrary neighbourhood, note that we can do the following: Let $y\in\mathbb{R}^k$ and $\epsilon>0$. Set a rational number $x_i$ such that $y_i - \epsilon < x_i < y_i + \epsilon$, with $x=(x_1,...,x_k)$. Then we have \begin{equation*} d(x,y)<\sqrt{\sum^{k}_{i=1}\epsilon^2}=\sqrt{k}\epsilon\to 0, \end{equation*} which completes the proof.