Two subsequences converge. Does a third one converge? The sequence $(w_n)_{n \in \mathbb{N}}$ has the property
$$\forall n \in \mathbb{N}: |w_{n+1}-w_n|<{1 \over n}$$
Assume $A<B<C$ and $A, C$ are the limits of two subsequences of $(w_n)$. How can we show that there is a subsequence $(w_{n_k})$ that converges to $B$.
 A: We will show that $B$ is an accumulation point of $(w_n)_n$, i.e. that:
$$ \forall\varepsilon> 0, \forall N \geq 0 \exists n \geq N \text{ s.t. } \lvert w_n - B \rvert < \varepsilon
$$

Fix $\varepsilon > 0$. By assumption,
$$\forall N \geq 0 \exists n \geq N \text{ s.t. } \lvert w_n - A \rvert < \varepsilon \tag{1}
$$
and
$$\forall N \geq 0 \exists n \geq N \text{ s.t. } \lvert w_n - C \rvert < \varepsilon \tag{2}
$$
Fix any $N\geq0$, and $M \stackrel{\rm def}{=} \max(N,1/\varepsilon)$, and let $n,m\geq M$ be as guaranteed respectively by (1) and (2) above for $M$, i.e. such that $\lvert w_n - A \rvert < \varepsilon$ and $\lvert w_m - C \rvert < \varepsilon$.
Without loss of generality, assume $n<m$. Note that $\lvert w_n - w_m\rvert < C-A + 2\varepsilon$ by the triangle inequality.
If $\lvert w_n - B\rvert < \varepsilon$ or $\lvert w_m - B\rvert < \varepsilon$, we are done. Assume otherwise.
Since $w_m - w_n = \sum_{k=0}^{m-n-1} (w_{n+k+1}-w_{n+k})$ with $w_n < B-\varepsilon$, $w_m > B + \varepsilon$; and each summand of the RHS is at most $\frac{1}{M} \leq \varepsilon$ by assumption, there must exist $0\leq k\leq m-n$ sch that $\lvert w_{n+k} - B \rvert \leq \varepsilon$.
This shows that 
$$\forall N \geq 0 \exists n \geq N \text{ s.t. } \lvert w_n - B \rvert < \varepsilon \tag{3}
$$
and since $\varepsilon>0$ was arbitrary, that $B$ is a limit point of $(w_n)_n$.
