Every open covering of a subset of $\Bbb R^n$ has a countable subcover I want to show the title. More formally,
Let $X\subset\Bbb R^n$ be a subset and $\{U_\alpha\}_{\alpha\in I}$ be a open covering of $X$. Then there is a subcover $\{U_\alpha\}_{\alpha\in I'\subset I}$ that covers $X$.
My basic plan is that define $S:=\Bbb Q^n$. Then for each $r\in S$, if there is $U_r\in\{U_\alpha\}_{\alpha\in I}$ that contains $r$ then choose only one such $U_r$. Collect them all for each $r$ (if there's such). I want to show this $\{U_r\}$ covers $X$. Any help? or any other idea?
 A: Your idea will, unfortunately, not quite work. Take for instance $X = \mathbb R$. Enumerate the rationals $\mathbb Q$ as $q_n$ and let $U_n = (q_n - 2^{-n}, q_n + 2^{-n})$. Take also $U_* = \mathbb R$ itself. Now, your construction is to take exactly one open neighborhood of every rational point. This won't work for any arbitrary choice. For instance, each $q_n \in U_n$ but the union $\bigcup U_n$ is not all of $\mathbb R$. The sum of the lengths of all the $U_n$ is $\sum 2^{-n} = 2$, so they cannot cover $\mathbb R$ even though we have a neighborhood of every rational point.
But as I said, your intuition to use $\mathbb Q$ is good. Recall that the definition of an open subset $U$ of $\mathbb R$ is that for all $x \in U$ there is a ball containing $x$ that is in $U$. Symbolically, $x \in B \subseteq U$. Such a ball is determined by two parameters - its center $p \in \mathbb R^n$ and its radius $r > 0$, denoted $B = B_p(r)$. We can perturb this a bit to show that $U$ is open iff every point $x \in U$ is contained in a rational ball $B \subseteq U$. We say that $B = B_p(r)$ is rational if $p \in \mathbb Q^n$ and $r \in \mathbb Q$. Importantly, there are only countably many rational balls!
Here's the proof. If $U$ is covered by rational balls then it's in particular covered by balls and is therefore open. Suppose conversely that $U$ is open and take $x \in U$. Then by assumption there is a ball $x \in B_p(r) \subseteq U$. By density, we can find rational points arbitrarily close to $x$. Then indeed we can find a rational point $q \in \mathbb Q^n$ such that $x \in B_q(\varepsilon) \subseteq B_p(r)$ Imagine this as nudging $p$ ever so slightly to a very nearby rational point $q$. We can additionally find some rational $s > \varepsilon$ that is just ever so slightly larger, so that $B_q(s) \subseteq B_p(r)$. Then $x \in B_q(s) \subseteq U$ is a rational ball. I didn't do this in great detail, but if you draw a picture you should be able to prove this more rigorously.
Anyways, take now your open cover $\{U_\alpha\}_{\alpha \in I}$. Then for each $x \in X$ there is some rational ball $x \in U_{\alpha}$. By the above, we can find a rational ball $x \in B^\alpha \subseteq U_{\alpha}$. There are only countably many rational balls $B^\alpha$, and for each ball pick exactly one $U_\alpha$ containing it. Then we have chosen countably many elements of the cover. Furthermore, we have that each $x$ is in such a rational ball. Hence, these chosen $U_\alpha$ are a countable subcover.
A: $\mathbb R^n $ is second countable:  the open balls with centers with rational coordinates and rational radii form a basis for the topology.
The property you are referring to is called the Lindelöf property.   Any second countable space is Lindelöf.
