The catch in proving that a singleton is a Borel set? Let $\Omega= (0,1]$. Let $\mathcal{C}$ be the collection of all open interval in $(0,1]$.
By open interval in $(0,1]$ I mean any interval of the form $(a,b)$, where $a\geq 0$ and $b\leq 1$.
Let $\sigma(\mathcal{C})$ be the smallest sigma-algebra that contains $\mathcal{C}$. We call this sigma-algebra a Borel-sigma algebra. Now, let us we have to prove that the singleton $\{0.7\}$ is in $\sigma(\mathcal{C})$. In order to prove it, we use the following result:
\begin{equation}
\{0.7\}=\bigcap_{n=1}^{\infty}\Big[(0.7-1/n, 0.7+1/n)\cap(0,1]\Big].
\end{equation}
The above result is based on the fact that
\begin{equation}
(0.7-1/n, 0.7+1/n)\cap(0,1]\in\mathcal{C}\in\sigma(\mathcal{C}), \mbox{for}\ n=2,3,4...
\end{equation}
Therefore the intersection of $\bigcap_{n=1}^{\infty}\Big[(0.7-1/n, 0.7+1/n)\cap(0,1]\Big]$ would be contained in $\sigma(\mathcal{C})$.
However, here is the catch, for $n=2$, we have
\begin{equation}
(0.7-1/n, 0.7+1/n)\cap(0,1]= (0.2, 1.2)\cap(0,1]=(0.2, 1]\notin \mathcal{C}
\end{equation}
Therefore the quantity $\bigcap_{n=1}^{\infty}\Big[(0.7-1/n, 0.7+1/n)\cap(0,1]\Big]$ does not have to be contained in $\sigma(\mathcal{C})$.
Now, can anybody help me to figure out how my argument is not valid?
 A: Understanding the topology (this part is to address the confusion in the comments under the original question): Let us first make sure we have the right notion of open intervals in $\Sigma = \left(0,1\right]$. Let us assume that $\mathbb{R}$ is endowed with the standard topology $\mathcal{T}_{\mathbb{R}}$ generated by open intervals, which are in $\mathbb{R}$ defined as $(a,b)$, with $a < b$ and $a,b \in \mathbb{R}$. This also seems like the topology you would like to work with and thus it seems like a natural choice. Let us assume furtermore that $\Omega = \left(0,1 \right] \subset \mathbb{R}$ is endowed with the subspace topology defined as
$$\mathcal{T}_{\mathbb{\Omega}} := \{\Omega \cap U: U \in \mathcal{T}_{\mathbb{R}} \}.$$
As @311411 and @mathcounterexamples.net already pointed out, $\left(0,1\right]$ is an open interval in $\left(0,1\right]$ since it lies in the subspace topology and since it is an interval. Even more so: any interval of the form $\left(a,1\right]$ with $0<a<1$ is open in $\left(0,1\right]$ since they can be written as $\left(a,1\right] = \Omega \cap (0,2)$ and since  $(0,2) \in \mathcal{T}_{\mathbb{R}}$. However, any interval of the form $\left(a,b\right]$ with $0<a<b<1$ is not open since it is not open in $\mathbb{R}$ and since it can be directly compared to open sets in $\mathbb{R}$, namely $\Omega\cap \left(a,b\right] = \left(a,b\right]$ (likewise with the square- and standard bracket reversed). This means that all the other intervals in $\left(0,1\right]$ can only be called open if they are of the form $(a,b)$ with $0 < a < b < 1$, just as you are used to in $\mathbb{R}$ (check for yourself that this is indeed the case by checking that they lie in the subspace topology).
Now proving that a singleton is a Borel set: Once you have established the right notion of a topology on $\left(0,1\right]$ and given that we work with the standard definition of intervals, we can now have a understanding of this collection $\mathcal{C}$: the intervals $(a,b)$ with $0<a<b<1$ and the intervals $\left(a,1\right]$ with $0<a<1$. The Borel $\sigma$-algebra over $\left(0,1\right]$ can now indeed be defined as the $\sigma$-algebra generated by $\mathcal{C}$.
Then now showing that any singleton $\{x\}$ lies in $\sigma(\mathcal{C})$ comes down to showing that we can find the right (infinite) combination of sets in $\sigma(\mathcal{C})$ that generate $\{x\}$, according to the definition of $\sigma$-algebras. As @Kurt G pointed out, we can always find $n_{0}$ such that for any $n \geq n_0$ we have that
$$\left(x - \frac{1}{n}, x + \frac{1}{n}\right) \in \sigma(\mathcal{C}),$$
namely:
$$n_0 = \lceil \frac{1}{\max\{ \frac{x}{2}, \frac{1-x}{2} \}} \rceil,$$
(this is really just solving inequalities, I leave this to the OP for now).
Now by definition of the $\sigma$-algebra we can find that
$$\{x\} = \bigcap_{n\geq n_0} \left(x - \frac{1}{n}, x + \frac{1}{n}\right) \in \sigma(\mathcal{C}).$$
If you see any mistakes, please let me know.
