n-dimensional Euclidean space is separable

Question

I'm self-teaching topology, and currently, I'm working on the topic of separable space. The following sentences are what I saw in Wikipedia:

1. Every compact metric space (or metrizable space) is separable.
2. Any topological space that is the union of a countable number of separable subspaces is separable. Together, these first two examples give a different proof that n-dimensional Euclidean space is separable.

The second statement makes me feel confused. How should I prove the union of a countable number of separable subspaces is separable, and why combining 1 and 2 indicates that n-dimensional Euclidean space is separable?

Answer 1

To answer your second question first, note that the interval $I_k=[k,k+1]$ is compact for each $k\in\Bbb Z$. Let $\langle a_1,\ldots,a_n\rangle\in\Bbb Z^n$; then $\prod_{k=1}^nI_{a_k}$ is a compact metric space, so by (1) it is separable. Finally, $\Bbb Z^n$ is countable, and

$$\Bbb R^n=\bigcup\left\{\prod_{k=1}^nI_{a_k}:\langle a_1,\ldots,a_n\rangle\in\Bbb Z^n\right\}\,,$$

so (2) implies that $\Bbb R^n$ is separable.

To prove (2), let $X=\bigcup_{n\in\Bbb N}X_n$, where each $X_n$ is separable. This means that each $X_n$ has a countable dense subset $D_n$. Let $D=\bigcup_{n\in\Bbb N}D_n$; $D$ is the union of countably many countable sets, so $D$ is countable. Finally, let $U$ be any non-empty open subset of $X$; there is at least one $n\in\Bbb N$ such that $U\cap X_n\ne\varnothing$. And $U\cap X_n$ is a non-empty open subset of $X_n$, so $U\cap D_n\ne\varnothing$: $D_n$ is dense in $X_n$, so it intersects every non-empty open subset of $X_n$. But then $U\cap D\supseteq U\cap D_n\ne\varnothing$, so we’ve shown that every non-empty open subset of $X$ intersects $D$, i.e., that $D$ is dense in $X$. Thus, $X$ has a countable dense subset and is therefore separable.