First, $\Bbb{N}$ is the set of positive integers, so it is bounded below by zero. This means $\min{\Bbb{N}}=0$ or $1$, depending on whether or not you allow $0$ in the natural numbers. Your intuition on your first question is almost right, but you should find $$\bigcup_{t \in \Bbb{N}} \{x \mid t^2 \leq x \leq (t+1)^2 \} = [1,\infty)$$ because $\infty \notin \Bbb{R}$ as $\infty$ is not a number, and does not belong in the set $\Bbb{R}$ of all real numbers. The half open interval $[1,\infty)$ makes that clear. On your second problem, the answer should be $(-\infty, \infty)$ for similar reasons. For all $y\in \Bbb{R}$ you can find $t \in \Bbb{N}$ such that $|y| \leq t$, and hence $y \in [-t,t]$, which means $$\Bbb{R} \subseteq \bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \}$$ The reverse inclusion follows easily since a union of subsets of $\Bbb{R}$ is itself a subset of $\Bbb{R}$, so you have $$\bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \} = \Bbb{R} = (-\infty, \infty)$$
The solution for $(c)$ is perfect up to $\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$, so let's keep going from here.
It is easy to lose track of what we are doing with these kind of problems, so let's not forget that we are trying to approximate the set $E$ from outside, and that right now we only have an approximation for every $n$. The obvious thing to do is then to consider the union of the approximating sets: let $$C_{\epsilon} = \bigcup_{n = 1}^{\infty}C_n.$$
The $\epsilon$ in $C_{\epsilon}$ is there to stress the dependence upon $\epsilon$ of the $C_n$'s and hence of $C$. It is worth noticing that $C_{\epsilon} \in \mathcal{A}_{\sigma}$ being the countable union of countable unions of elements of $\mathcal{A}$.
For the next step to work you might want to state clearly in your proof that we can take $C_n \subset \cup_{i = 1}^nX_i$.
Intuitively, $C_{\epsilon}$ is a decent approximation of $E$. Let's make this precise:
\begin{align}
\mu^*(C_{\epsilon} \cap E^c) = & \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E^c)\Big) = \mu^*\Big(\bigcup_{n = 1}^{\infty} (C_n \cap E_n^c)\Big) \\ \le & \sum_{n = 1}^{\infty}\mu^*(C_n \setminus E_n) \le \epsilon.
\end{align}
(good job considering the $2^{-n}$, that really came in handy!)
Let's go back to our intuition: $C_{\epsilon}$ is an $\epsilon$-good approximation of $E$, therefore we would like to send $\epsilon$ to $0$ to get the best possible approximation. Since we want to end up in $\mathcal{A}_{\sigma \delta}$ what we need to do is to "discretize" the sets $C_{\epsilon}$ considering instead the sets $C_{\frac{1}{n}}$. (the notation is unfortunate, these resemble too much the $C_n$ sets, hopefully this won't cause any confusion)
Then clearly $$C := \bigcap_{n = 1}^{\infty}C_{\frac{1}{n}} \in \mathcal{A}_{\sigma \delta}$$ and satisfies $\mu^*(C \setminus E) = 0$, proving the claim.
To show this last equality notice that $$\mu^*(C \setminus E) \le \mu^*(C_{\frac{1}{n}} \setminus E) \le \frac{1}{n} \to 0.$$
Let me know if there is anything that I need to clarify!
Best Answer
Actually, none of those options are correct, since $\Bbb N$ is unbounded neither $\max\Bbb N$ nor $\sup\Bbb N$ exist. In fact,
$$\bigcap_{n\in\Bbb N}[n,\infty)=\emptyset.$$
The notation $(\infty,\infty)$ can be considered correct I suppose, but $(\infty,\infty)=\emptyset$.