I know that every open set in $\mathbb R$ is a countable union of open intervals, c.f. Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]. However, I'm not so sure what would be a good way to prove this in $\overline{\mathbb R}$, the extended real line. According to Rudin's Real and Complex Analysis, the topology $\tau$ in $\overline{\mathbb R}$ consists of sets of the form $(a,b), [-\infty, a), (a, \infty]$ and any union of segments of this type, where $a,b \in \mathbb R$ are arbitrary real numbers. I list my proof below, and would appreciate it if someone can confirm its validity, or point out where I'm mistaken, or provide a better proof.
My Attempted Proof: Suppose $V\in \tau$, i.e. $V=\bigcup_{t\in S} A_t$ is an open set in $\overline{\mathbb R}$, where $A_t$ is a segment of the form $(a,b), [-\infty, a), (a, \infty]$. If $-\infty\notin V$ and $\infty \notin V$, then the proof is the same as that in $\mathbb R$.
Otherwise, suppose $-\infty\in V$. Then some $A_t$ must be of the form $[-\infty, a_t)$. Let $S_0$ be the set of all such $t$'s, and $V_0=\bigcup_{t \in S_0}A_t$. It follows that $V_0=[-\infty, a_0)$, where $a_0=\sup_{t\in S_0} a_t$. By the same token, if $\infty \in V$, then let $S_1=\{t:A_t=(a_t, \infty]\}$ and $V_1=\bigcup_{t \in S_1}A_t=(a_1, \infty]$, where $a_1=\inf_{t\in S_1} a_t$.
Now let $S_2=S\setminus(S_0\cup S_1)$ and $V_2=\bigcup_{t \in S_2}A_t$. Note that $V_2$ is an open set in $\mathbb R$, so it must be an at most countable union of segments of the form $(a,b)$. The proof is hence completed by noting that $V=V_0\cup V_1\cup V_2,$ so it must be an at most countable union of segments of the form $(a,b)$, $[-\infty, a),$ or $(a, \infty]$.
Is this proof correct? Are there better/simpler ways to prove this? Thanks a lot!
