Completion of a separable metric space is separable

Question

One of my textbooks states that the completion of a metric space is separable when the metric space is separable. I am not sure why that is the case. Could someone explain this to me?

Answer 1

Because the original space is densely embedded in its completion. Therefore the countable dense subset of the original space is dense in the completion.

Added: Let $\iota: X\to \overline X$ be the embedding an $\operatorname{cl} C=X$ with $C$ countable. Every non-empty open set $U\subseteq\overline X$ intersects $\iota[X]$. Hence $\iota^{-1}[U]$ is non-empty and open, and therefore $\iota^{-1}[U]$ intersects $C$. This means that $\iota[C]$ intersects $U$, establishing density of the countable set $\iota[C]$.