Given $\Omega$ a nonempty set and $C \subset \Omega$ is there a general way to find $\sigma(C)$ (the smallest $\sigma$-algebra containing $C$) Is there a general way to solve the problems where they ask for the smallest $\sigma$-algebra containing a subset?
I'm trying to solve the following problem:

Given $C=\{\{x\}:x\in \mathbb{R}\}$ find $\sigma(C)$ and prove that
$\sigma(C)\subset \mathcal{B}$ but $\sigma(C)\neq \mathcal B$

The first thing that has come to my mind is that as in order to build a $\sigma$-algebra containing $C$ this has to have the properties of a regular $\sigma$-algebra, then for any given familly of subsets of $\{A_n\}_{n=1}^{\infty}$  their union has to be an element of the $\sigma$-algebra. Therefore, as the $\sigma$-algebra we are looking for contains $C$ by definition, for any familly of subsets of $C$ this propperty has to hold aswell. Then, for any given subset $S$ of $\mathbb R$ we can find a family of $\sigma(C)$ such that $S=\cup_n A_n$ and therefore $\sigma(C) = \mathcal P(\mathbb R)$.
Now, If this holds, I can't find a way to prove the last part of proving how this $\sigma$-algebra is contained in Borel's $\sigma$-algebra but its not equal to it.
I'd appreciate someone to point out where I'm wrong on the first part or to give me a tip on how to do the second one.
I also feel like if this were to be another subset I wouldn't be able to figure out the smallest $\sigma$-algebra containing it. Is there a general way to face these kind of problems?
Thanks in advance
 A: 
Then, for any given subset $S$ of $\mathbb R$ we can find a family of $\sigma(C)$ such that $S=\cup_n A_n$ and therefore $\sigma(C) = \mathcal P(\mathbb R)$.

That makes no sense. To build a $\sigma$-algebra out of a set $C$ you need for sure to contain countable unions and complements of subsets. But you cannot write an arbitrary set $S$ (an interval, for example) as a countable union of finite sets, so your assertion above is wanting.
The $\sigma$-algebra $\sigma(C)$ is necessarily contained in $\mathcal B$, because $C$ is (points are closed, so they are Borel). As mentioned, $\sigma(C)$ will contain all countable sets (countable unions of points) and their complements. Let us try the following: let
$$
C_\sigma=\{S\subset\mathbb R:\ S\ \text{ or }\ S^c \text{ is countable }\}.
$$
We have for sure that $C_\sigma\subset\sigma(C)$. So if $C_\sigma$ is a $\sigma$-algebra, then we are done. I'll leave this last part to you.
Finally, the inclusion $\sigma(C)\subset\mathcal B$ is proper; to see this, show that intervals like $(0,1)$ cannot possibly be in $\sigma(C)$.
A: It's not correct that the union of any family of subsets of $C$ has to be in $\sigma(C)$.  That's only true for countable collections of elements of $C$.  To prove that $\sigma(C) \subseteq \mathcal B$, it's enough to prove $\forall x \in \Bbb R~(\{ x \} \in \mathcal B)$.
