How to prove for every open cover of the real numbers $\mathbb{R}$ there is a countable subcover? Without using more sophisticated results from topology, assuming only a real analysis background.
I've found a proof using second-countable space characterization, but since i never studied general topology before, it's hard to associate a countable base on the real line. My intuition says to transform the open cover into disjoint open subsets, but how to achieve that?
I’m going to assume that you want to prove that $\mathbb{R}$ is Lindelöf. You definitely do not want to try to transform the open cover into disjoint open sets, because it can’t be done: no family of two or more pairwise disjoint non-empty open sets covers $\mathbb{R}$.
Getting a countable base for $\mathbb{R}$ isn’t at all hard, provided that you know that $\mathbb{Q}$, the set of rational numbers, is countable. Just let $\mathscr{B}$ be the set of open intervals with rational endpoints: each pair $\{p,q\}$ of distinct rational numbers determines exactly one such interval, $(p,q)$ if $p<q$, and $(q,p)$ if $p>q$, and there are only countably many pairs of rational numbers, so $\mathscr{B}$ is countable. It only remains to show that $\mathscr{B}$ is a base for the topology of $\mathbb{R}$, which just means showing that every open set in $\mathbb{R}$ is a union of members of $\mathscr{B}$.
Every non-empty open set in $\mathbb{R}$ is a union of open intervals. If we can show that every open interval in $\mathbb{R}$ is a union of members of $\mathscr{B}$, i.e., of open intervals with rational endpoints, it will immediately follow that every non-empty open subset of $\mathbb{R}$ is also such a union. To this end let $(a,b)$ be any non-empty open interval in $\mathbb{R}$. Then there are sequences $\langle p_n:n\in\mathbb{N}\rangle$ and $\langle q_n:n\in\mathbb{N}\rangle$ of rational numbers such that:
In other words, $\langle p_n:n\in\mathbb{N}\rangle$ is a decreasing sequence converging to $a$, $\langle q_n:n\in\mathbb{N}\rangle$ is an increasing sequence converging to $b$, and $p_0<q_0$. It easily follows that $$(a,b) = \bigcup_{n\ge 0}(p_n,q_n)\;,$$ and each interval $(p_n,q_n)$ obviously has rational endpoints. Thus, every non-empty open interval in $\mathbb{R}$ is a union of members of $\mathscr{B}$, so every non-empty open set of any kind in $\mathbb{R}$ is such a union, and $\mathscr{B}$ is therefore a countable base for the topology of $\mathbb{R}$.
It’s now trivial to see that $\mathbb{R}$ is Lindelöf: if $\mathscr{U}$ is any open cover of $\mathbb{R}$, just let $$\mathscr{B}_\mathscr{U}=\{B\in\mathscr{B}:\exists U\in\mathscr{U}\big(B\subseteq U\big)\}.$$ Each $U\in\mathscr{U}$ is the union of the members of $\mathscr{B}_\mathscr{U}$ contained in it, so $\mathscr{B}_\mathscr{U}$ covers $\mathbb{R}$. It’s also countable, since it’s a subset of the countable set $\mathscr{B}$. Now for each $B\in\mathscr{B}_\mathscr{U}$ choose some $U(B)\in\mathscr{U}$ such that $B\subseteq U$; the definition of $\mathscr{B}_\mathscr{U}$ guarantees that there is one. Let $\mathscr{U}_0=\{U(B):B\in\mathscr{B}_\mathscr{U}\}$. Then $\mathscr{U}_0$ is countable, since it’s no bigger than $\mathscr{B}_\mathscr{U}$, and $$\bigcup\mathscr{U}_0 = \bigcup_{B\in\mathscr{B}_\mathscr{U}}U(B)\supseteq \bigcup_{B\in\mathscr{B}_\mathscr{U}}B=\mathbb{R},$$ so $\mathscr{U}_0$ is indeed a countable subcover of $\mathscr{U}$.
Added: This idea can be extended to $\mathbb{R}^n$. Instead of open intervals with rational endpoints, you take for your countable base the set of Cartesian products of such intervals. In other words, you take as a base for $\mathbb{R}^n$ the set of open boxes of the form $B_1\times\dots\times B_n$, where $B_1,\dots,B_n\in\mathscr{B}$. Then you show that each open box in $\mathbb{R}^n$ is a union of these ‘rational boxes’. Since there are only countably many rational boxes, and every non-empty open set in $\mathbb{R}^n$ is a union of them, it follows that $\mathbb{R}^n$ is Lindelöf: the remainder of the argument is just like that for $\mathbb{R}$.
There is another way to proceed, if you know that closed, bounded subsets of $\mathbb{R}^n$ are compact, meaning that every open cover of such a set has a finite subcover. I’ll do it for $\mathbb{R}$; the generalization to $\mathbb{R}^n$ is pretty straightforward. $\mathbb{R}$ is the union of the closed intervals $[n,n+1]$ for $n\in\mathbb{Z}$. There are only countably many such intervals, and each of them is compact. Now let $\mathscr{U}$ be any open cover of $\mathbb{R}$. For each $n\in\mathbb{Z}$ let $\mathscr{U}_n = \{U\in\mathscr{U}:U\cap [n,n+1]\ne\varnothing\}$. Then $\mathscr{U}_n$ is an open cover of $[n,n+1]$, so it has a finite subcover, $\mathscr{V}_n$. Finally, let $$\mathscr{V}=\bigcup_{n\in\mathbb{Z}}\mathscr{V}_n\;;$$ $\mathscr{V}$ is the union of countably many finite sets, so it’s a countable subset of $\mathscr{U}$, and it clearly covers $\mathbb{R}$.
NOTE: I don't know if this is what the user would call "sophisticated results". It uses some ostensibly highfalutin language, but the ideas are very simple. I hope it suffices.
I think what the user is asking is to prove that every second countable space is Lindelof (in more common notation).
Really, the user is asking to prove that "If $X$ is second countable and $A$ is a subset of $X$, the any open cover of $A$ admits a countable subcover." To see how $\text{Second Countable}\implies \text{Lindelof}$ gives us this merely note that if $\Omega$ is a $X$-open cover of $A$ then $\Omega$ induces a $A$-open cover of $A$ and since $A$ is second countable (since second countability is hereditary) our result $\text{Second Countable}\implies\text{Lindelof}$ gives us what we want.
So, let's prove $\text{Second Countable}\implies\text{Lindelof}$. So, let $X$ be second countable with countable basis $\mathscr{B}$, and let $\Omega=\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an open cover for $X$. By assumption, for each $\alpha\in\mathcal{A}$ we can cover $U_\alpha$ with some collection $B_\alpha$ of elements of $\mathscr{B}$. Note then that $\displaystyle \Sigma=\bigcup_{\alpha\in\mathcal{A}}B_\alpha$ is a countable open cover for $X$. So, for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it. Then, this subset, call it $\Gamma$, of $\Omega$ is an open cover of $X$ (since its union contains the union over all the elements of $\Sigma$ which is $X$) and is countable since there is a surjection $\Sigma\to\Gamma$ and $\Sigma$ is countable. Thus, $\Gamma$ is our desired countable subcover of $\Omega$.
