$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I spent a considerable amount of time ($>2$ hours) thinking about the below questions, however my arguments are sloppy. Could you give me some hints on how to proceed with (b),(c) and (d)?
Decide whether the following statements are true or false. Provide counterexamples for those that are false, and supply proofs for those that are true.
(a) An open set that contains every rational number must necessarily be all of $\mathbf{R}$.
(b) The Nested Interval property remains true if the term "closed interval" is replaced by "closed set".
(c) Every nonempty open set contains a rational number.
(d) Every bounded infinite closed set contains a rational number.
(e) The Cantor Set is closed.
My Attempt.
(a) I think that this proposition is true. We proceed by contradiction. Let $O \supseteq \mathbf{Q}$ be an open set containing all the rationals. Assume that this is not the whole of $\mathbf{R}$. Then, $\exists x \in O^C \cap \mathbf{R}$. Since $O^C$ is a closed set, there exists a sequence of rational numbers $(q_n) \in O^C$, such that $\lim q_n = x$, with $q_n \ne x$ for all $n \in\mathbf{N}$. But, this implies $O \not\supseteq Q$. This leads to a contradiction. Hence, our initial assumption is false.
(b) I could not think of counterexamples here. For instance, consider the decreasing sequence of closed sets
\begin{align*}
A_n = \left\{-\frac{1}{n},\frac{1}{n}\right\}
\end{align*}
We have $A_1 \supseteq A_2 \supseteq A_3 \ldots$ and $\cap_{n=1}^{\infty} A_n = \{0\} \ne \emptyset$.
Another example could be,
\begin{align*}
A_n = \left\{-1-\frac{1}{n},1+\frac{1}{n}\right\}
\end{align*}
(c) Let $O$ be a non-empty open set and say, $x \in O$. Then, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$ such that $V_\epsilon(x) \subseteq O$. Since, $(x-\epsilon,x] \subseteq O$, and the set of real numbers $\mathbf{R}$ are dense in $\mathbf{Q}$, there exists a rational number $m/n$ between the two reals $x – \epsilon < \frac{m}{n} < x$. So, every non-empty open set $O$ contains a rational number.
(d) Consider the bounded infinite closed sets
\begin{align*}
A_n = \left\{1 + \frac{1}{n}:n \in \mathbf{N}\right\} \cup \{1\}
\end{align*}
\begin{align*}
B_n = \left\{\frac{1}{(\sqrt{2})^{2n-1}}:n \in \mathbf{N}\right\} \cup \{0\}
\end{align*}
\begin{align*}
C_n = \left\{x_n: x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}, x_1 = 1\right)\right\} \cup \{\sqrt{2}\}
\end{align*}
$A_n, B_n, C_n$ contain rational numbers. I couldn't think of a counterexample here.
By AoC, every bounded infinite closed subset of $\mathbf{R}$ has a least upper bound. Let $s = \sup A$. Since, $A$ is closed, $s \in A$. We have a dichotomy: either (i) the supremum is irrational or (ii) the supremum is rational number. If $s$ is rational, then we are done. If $s$ is irrational and then there exists a sequence of rationals in $A$, such that $\absval{q – s} < \epsilon$.
(e) The Cantor set is the limit set formed by recursively removing the open middle one-thirds. So, \begin{align*}
C = [0,1] – \left[\left(\frac{1}{3},\frac{2}{3}\right) \cup \left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right) \ldots \right]
\end{align*}
The countable union of open sets is open. Since $C$ is the complement of an open set, it is closed.
Best Answer
(a) This is false: $\Bbb R\setminus\left\{\pi\right\}\ne\Bbb R$.
(b) It is also false: $\displaystyle\bigcap_{n\in\Bbb N}[n,\infty)=\emptyset$.
(c) True, since $\Bbb Q$ is dense in $\Bbb R$. Indeed, that's what “dense” means.
(d) It is false. Take, say, $\displaystyle\{\pi\}\cup\left\{\pi+\frac1n\,\middle|\,n\in\Bbb N\right\}$.
(e) Your answer is correct.