Solutions

(a) Assume, for contradiction, that the natural numbers are bounded above. Let's also assume that this upper bound is $u$. Since the natural numbers are also increasing, we then know by MCT that the sequence of natural numbers converges.

Assume that this limit is $L$. By our definition of convergence, we have that for any $\epsilon > 0$,

$$\exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N \implies \abs{n - L} < \epsilon.$$

However, if we choose $\epsilon = 1$, the application of the definition implies that

$$\exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N \implies \abs{n - L} < 1.$$

which is impossible since $( n + 1 ) \not\in V_{1}(L)$. Therefore, the natural numbers are unbounded and

$$\forall x \in \mathbb{R}, \; \exists n \in \mathbb{N}: \quad x < n.$$

(b) Consider the intervals given by $I_{n} = [a_{n}, b_{n}]$, for all $n \in \mathbb{N}$, such that

$$I_1 \supseteq I_2 \supseteq I_3 \dots.$$

Notice that the sequence $a_{n}$ is increasing (a contradiction would arise if it were decreasing) and, likewise, the sequence $b_{n}$ is decreasing. Since $a_{n}$ is bounded above by $b_1$ and $b_{n}$ is bounded below by $a_1$, then both sequences converge.
Now, assuming that $a_{n}$ converges to $a$ and that $b_{n}$ converges to $b$, we can apply the Order Limit Theorem. Thus,

$$\forall n \in \mathbb{N}, \quad a_{n} \leq b_{n} \implies a \leq b.$$

Therefore,

$$\bigcup_{n \in \mathbb{N}} I_{n} = [a, b] \neq \emptyset.$$