Here's a suggestion. By definition $\sigma(\mathcal{E})$ is the smallest sigma algebra containing $\mathcal{E}$. Thus if we want to show that $\newcommand\calA{\mathcal{A}}\newcommand\calB{\mathcal{B}}\newcommand\calE{\mathcal{E}}\sigma(\calE)=\calA$, a common strategy is to show that $\calA$ is a $\sigma$-algebra and that $\calE\subseteq \calA$ and if $\calB$ is any $\sigma$-algebra containing $\calE$ then $\calA\subseteq \calB$.
In your case, $\calA$ will be the set of all subsets $A$ of $\Omega$ such that either $A$ is countable or its complement is countable. Then to show $\sigma(\calE)=\calA$ you need to show:
- $\calA$ is a $\sigma$-algebra,
- $\calE\subseteq \calA$, and
- if $\calB$ is a $\sigma$-algebra containing $\calE$ then $\calA\subseteq \calB$.
Can you show these things?
Side note
In case its unclear why these properties show that $\sigma(\calE)=\calA$, note that the third property immediately gives us $\calA\subseteq \sigma(\calE)$ (since $\sigma(\calE)$ is a $\sigma$-algebra containing $\calE$), but we also have by definition that $\sigma(\calE)$ is the intersection of all $\sigma$-algebras containing $\calE$, which includes $\calA$ by properties 1 and 2, so $\sigma(\calE)\subseteq \calA$ as well.
Edit
Suggestions for part 3. Well I have a hint, but anything stronger than that is a solution, so here's the hint and the solution is in the spoiler tag.
Hint: How can a countable set be written in terms of elements of $\calE$?
Let $\calB$ be a sigma algebra containing $\calE$. First I claim any countable subset of $\Omega$ is in $\calB$, since if $A\subset\Omega$ is countable, then $A=\bigcup_{x\in A} \{x\}$, which is a countable union of elements of $\calE$, and therefore in $\calB$. If $A$ has a countable complement on the other hand, then its complement is in $\calB$ by the argument above, and so it is in $\calB$ as well. Thus $\calB$ contains $\calA$.
Each given cylinder set in $\mathcal C$ is of the form $f^{-1}(B)$ for some $B\in\mathcal B(\mathbb R^n)$, where $n\in\mathbb N$, $t_1,\ldots,t_n\in\mathbb T$ are a finite collection of indices, and the function $f:\mathbb R^{\mathbb T}\to\mathbb R^n$ is defined as $$f(x)\equiv(x_{t_1},\ldots,x_{t_n})\quad\text{for every $x\in\mathbb R^{\mathbb T}$}.$$ You want to show that $f^{-1}(B)\in\sigma(\mathcal E)$, meaning that $\mathcal C\subseteq\sigma(\mathcal E)$, which, in turn, will imply that $\sigma(\mathcal C)\subseteq\sigma(\mathcal E)$. (No need to use Dynkin’s theorem for this last implication—see below my remark on your proof of the reverse inclusion.)
To this end, define the following collection: $$\mathcal B^*\equiv\{B\subseteq\mathbb R^n\,|\,f^{-1}(B)\in\sigma(\mathcal E)\}.$$ It is not difficult to check that $\mathcal B^*$ is a $\sigma$-algebra on $\mathbb R^n$. Also, if $B=B_1\times\cdots\times B_n$ for some $B_1,\ldots,B_n\in\mathcal B(\mathbb R)$, then $f^{-1}(B)$ is an elementary cylinder by definition, so $f^{-1}(B)\in\mathcal E\subseteq\sigma(\mathcal E)$. Therefore, $$\{B_1\times\cdots\times B_n\,|\,B_1,\ldots,B_n\in\mathcal B(\mathbb R)\}\subseteq\mathcal B^*,$$ from which it follows that $$\mathcal B(\mathbb R^n)=\sigma\big(\{B_1\times\cdots\times B_n\,|\,B_1,\ldots,B_n\in\mathcal B(\mathbb R)\}\big)\subseteq\mathcal B^*,$$ since $\mathcal B^*$ is a $\sigma$-algebra. Therefore, if $B\in\mathcal B(\mathbb R^n)$, then $B\in\mathcal B^*$, so that $f^{-1}(B)\in\sigma(\mathcal E)$, which is precisely what you sought to show.
As for the first part of your proof, using Dynkin’s theorem is overkill. The fact that $\mathcal E\subseteq\mathcal C$ directly implies that $\mathcal E\subseteq\sigma(\mathcal C)$. Since $\sigma(\mathcal C)$ is a $\sigma$-algebra that includes $\mathcal E$, the smallest $\sigma$-algebra including $\mathcal E$ must be contained in $\sigma(\mathcal C)$. That is: $\sigma(\mathcal E)\subseteq\sigma(\mathcal C)$.
Best Answer
You do not need to show the other inclusion. Just remember that $\mathcal B = \sigma(\{\{x\}~|~x \in X\})$ is the smallest sigma-algebra that contains $\{\{x\}~|~x \in X\}$. The conclusion follows as, clearly, $\{\{x\}~|~x \in X\} \subset \mathcal{A} \subset \mathcal B$.