Proof that singleton sets are Borel sets in $\Omega = (0,1]$.

Question

I’m reading the following definition and lemma, and I’d like to ask some questions.

Definition 7.3
(a) Consider $\Omega=(0,1] .$ Let $\mathcal{C}_{0}$ be the collection of all open intervals in $(0,1] .$ Then $\sigma\left(\mathcal{C}_{0}\right)$, the $\sigma-$ algebra generated by $\mathcal{C}_{0}$, is called the Borel $\sigma$ - algebra. It is denoted by $\mathcal{B}((0,1])$.
(b) An element of $\mathcal{B}((0,1])$ is called a Borel-measurable set, or simply a Borel set.
Thus, every open interval in $(0,1]$ is a Borel set. We next prove that every singleton set in $(0,1]$ is a Borel set.

Lemma 7.4 Every singleton set $\{b\}, 0<b \leq 1$, is a Borel set, i.e., $\{b\} \in \mathcal{B}((0,1])$.

Proof: Consider the collection of sets set $\left\{\left(b-\frac{1}{n}, b+\frac{1}{n}\right), n \geq 1\right\} .$ By the definition of Borel sets. $$ \left(b-\frac{1}{n}, b+\frac{1}{n}\right) \in \mathcal{B}((0,1]) . $$ Using the properties of $\sigma$ -algebra, $$ \begin{aligned} &\left(b-\frac{1}{n}, b+\frac{1}{n}\right)^{c} \in \mathcal{B}((0,1]) \\ \Longrightarrow & \bigcup_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right)^{c} \in \mathcal{B}((0,1]) \\ \Longrightarrow &\left(\bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right)\right)^{c} \in \mathcal{B}((0,1]) \\ \Longrightarrow & \bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right) \in \mathcal{B}((0,1]) . \end{aligned} $$ Next, we claim that $$ \{b\}=\bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right) . $$ i.e., $b$ is the only element in $\bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right)$. We prove this by contradiction. Let $h$ be an element in $\bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right)$ other than $b$. For every such $h$, there exists a large enough $n_{0}$ such that $h \notin$ $\left(b-\frac{1}{n_{0}}, b+\frac{1}{n_{0}}\right) .$ This implies $h \notin \bigcap_{n=1}^{\infty}\left(b-\frac{1}{n}, b+\frac{1}{n}\right) .$ Using $(7.1)$ and $(7.2)$, thus, proves that $\{b\} \in$ $\mathcal{B}((0,1])$

1. I’m having a bit of trouble understanding what a typical element of $\sigma\left(\mathcal{C}_{0}\right)$ would be (or in general, what an element of a $\sigma$-algebra generated by a set is). Would it be correct to say that, if $F \in \sigma\left(\mathcal{C}_{0}\right)$, then either $F = \Omega \setminus S$, where $S$ is a countable union of open intervals in $\Omega$, or $F$ itself is a countable union of open intervals in $\Omega$?
2. In the proof, when checking that $\{b\} \in \mathcal{B}((0,1])$, are we only checking that the complement of $\{b\}$ is in $\mathcal{B}((0,1])$?
3. In the first sentence of the proof, for a fixed $b$, isn’t it true that not all $n$’s work?