Sequences whose terms are getting closer and closer

1. Let $l\in \R$ be an element outside $\mathcal{V}_{u}$, that is, there exist $\varepsilon >0$ and $N\in \N$ such that, for every $n\geq N$, we have $|u_{n}-l|>\varepsilon$. In other words, there exists an interval of radius $\varepsilon$ around $l$ into which no term of the sequence enters after some rank. But then, if we take for instance the interval of radius $\varepsilon/2$ around $l$, this interval is entirely contained in the complement of $\mathcal{V}_{u}$. Indeed, if $m$ belongs to this interval, then by the reverse triangle inequality, for $n\geq N$, we have:
   $$\left|m-u_n\right| \geqslant\left|\ell-u_n\right|-|\ell-m| \geqslant \frac{\varepsilon}{2}$$which shows that $m$ cannot belong to $\mathcal{V}_{u}$.

This shows that the complement of $\mathcal{V}_{u}$ is open in $\R$, and therefore $\mathcal{V}_{u}$ is closed.

2. Let $a$ and $b$ belong to $\mathcal{V}_{u}$, and without loss of generality assume that $a\leq b$. Let $c\in [a,b]$. We need to show that $c$ belongs to $\mathcal{V}_{u}$. Let therefore $\varepsilon >0$ and $N\in \N$. By assumption, there also exists $M\in \N$ such that, for every $n\geq M$, the sequence moves in small successive steps:
   $$|u_{n+1}-u_{n}|\leq \varepsilon.$$

Without loss of generality, assume that $N\geq M$ (otherwise, redefine $M=N$). Since $a$ is a subsequential limit, there exists $n\geq N$ such that
$$|u_{n}-a|\leq \varepsilon.$$Likewise, since $b$ is a subsequential limit, there exists $m\geq n$ (!) such that $|u_{m}-b|\leq \varepsilon$.

• If $a<c<u_{n}$, then we are done, since $\left|u_n-c\right|<\left|u_n-a\right| \leqslant \varepsilon$.
• If $u_{m}<c <b$, then we are done, since $\left|u_m-c\right| \leqslant\left|u_m-b\right| \leqslant \varepsilon$.
• Otherwise, we have $a<u_{n}<c<u_{m}<b$.
  But then, between the indices $n$ and $m$, since the sequence moves in small successive steps of size at most $\varepsilon$ and eventually exceeds $c$, there must necessarily exist an intermediate index $n_{0}$ between $n$ and $m$ such that $|u_{n_{0}}-c|\leq\varepsilon$. This is exactly what we wanted to prove!

For the second part of the question, consider the following sequence, given by the partial sums of the harmonic series:
$$u_{n}=\sum_{k=1}^{n}\frac{1}{k}.$$This is a classical example of a sequence for which the difference between successive terms tends to $0$, but which nevertheless diverges. Since the sequence is increasing and diverges, it has no cluster point. Therefore $\mathcal{V}_{u}$ is empty in this case.