Let $\textbf{S}$ and $\textbf{T}$ are two collections of subsets of $\mathbb{R}$, then $\sigma (\textbf{S})$ the smallest $\sigma-$ algebra that contains $\textbf{S}$ and $\sigma (\textbf{T})$ the smallest $\sigma-$ algebra that contains $\textbf{T}$.\
Distinguish between 2 cases:\
Case(I). $S \cap T = \emptyset.$\
Then, $S \cap T \in \sigma(\textbf{S}) \cap \sigma (\textbf{T}).$ Because each $\sigma-$algebra contains the empty set as the empty union equals the empty set and the $\sigma-$algebra is closed with respect to the formation of countable unions.\newline And hence $\sigma(S \cap T) \subset \sigma(\textbf{S}) \cap \sigma (\textbf{T})$.As $\sigma(S \cap T)$ is the smallest sigma algebra that contains $S \cap T.$Note that $\sigma(S \cap T)$ is the trivial $\sigma-$algebra in this case $i.e.\sigma(S \cap T) = \{\emptyset, \mathbb{R}\}$.
Case(II). $S \cap T \neq \emptyset.$\
Let $E \in S \cap T,$ then $E \in \sigma(\textbf{S})$ and $E \in \sigma(\textbf{T}).$ which means $ E \in \subset \sigma(\textbf{S}) \cap \sigma (\textbf{T}).$ $$i.e. S \cap T \subset \sigma(\textbf{S}) \cap \sigma (\textbf{T}). $$
But $\sigma(S \cap T)$ is the smallest sigma algebra that contains $S \cap T.$ Then $$\sigma(S \cap T) \subset \sigma(\textbf{S}) \cap \sigma (\textbf{T})$$
My question is:
I know that the last line in my argument in $Case(II)$ "But $\sigma(S \cap T)$ is the smallest sigma-algebra that contains $S \cap T.$ Then $$\sigma(S \cap T) \subset \sigma(\textbf{S}) \cap \sigma (\textbf{T})$$" is incorrect as an example the Borel sets and sigma-algebra of the open sets (I was given this example by a professor but I did not fully understand it, so could anyone explain this for me also?),Also, could anyone find the mistakes in my proof and give me an elegant and simpler proof than this, please?
From your comments, it sounds like your professor is critiquing your justification rather than saying your statement is incorrect. You either need to re-write your justification or simply justify the additional step of $$\sigma(\sigma(S) \cap \sigma(T)) = \sigma(S) \cap \sigma(T)$$
As for your request of a simpler proof... the acceptable justification for homework depends on what theorems have been proven for you to use, which is something I don't know. However, based on the professor's comments, it would seem you can at least use $$X\subseteq Y \implies \sigma(X) \subseteq \sigma(Y)$$
In this case, the typical proof of the statement you're proving would just be $$\left.\begin{array}{c}S \cap T \subseteq S \implies \sigma(S \cap T) \subseteq \sigma(S) \\ S \cap T \subseteq T \implies \sigma(S \cap T) \subseteq \sigma(T)\end{array}\quad\right\}\implies \sigma(S\cap T) \subseteq \sigma(S) \cap \sigma(T)$$