Convergence of subsequences of sequence satisfying $|d_n - d_{n-1}|<\frac{1}{n}$ Sequence $d_n$ satisfies $|d_n - d_{n-1}| < \frac{1}{n}$ for all $n \in N$. $a_n$ and $c_n$ are subsequences of $d_n$ such that $\lim_{n->\infty}a_n = A$, $\lim_{n->\infty}c_n = C$, $A < C$. Prove that for every $B$ that satisfies $A < B < C$ there is a subsequence $b_n$ of $d_n$ such that $\lim_{n->\infty} b_n = B$.
Intuitively, I can see that for the assumptions to happen, $d_n$ has to go from $A$ to $C$, then from $C$ to $A$ and so on, but I have no idea how to formalize it.
 A: Here is my attempt.
Call $\phi$ the subsequence for $a$, $\psi$ the subsequence for $c$. 
To show convergence of a subsequence towards $B$ amounts to show that for any $\epsilon > 0$ (take $\epsilon$ small enough so that the three sets $(A - \epsilon, A + \epsilon), (B-\epsilon, B+\epsilon)$ and $(C - \epsilon, C + \epsilon)$ have no intersection), any $N > 0$, you can find $n \geq N$ such that $d_n \in (B-\epsilon, B + \epsilon)$.
Consider thus $\epsilon > 0$, and $N$ large enough such that 


*

*$a_{\phi(n)} \in (A-\epsilon, A+\epsilon)$ for all $n \geq N$.

*$c_{\psi(n)} \in (C - \epsilon, C + \epsilon)$ for all $n \geq N$.

*$N > \frac{1}{2\epsilon}$.


Let $n \geq N$ and assume wlog $\phi(n) < \psi(n)$. Between $\phi(n)$ and $\psi(n)$, we have $\psi(n) - \phi(n)$ values to place in bands of width at most $\frac{1}{\phi(n)}$. So for $n$ large enough, we will have to place one into $(B-\epsilon, B+\epsilon)$. More precisely, let $l, l' \in \{\phi(n), \dots, \psi(n)\}$ such that 
$$ d_l < B - \epsilon, d_{l'} > B + \epsilon. $$
We have
$$|d_{l'} - d_l| \leq \frac{1}{l} < \frac{1}{\phi(n)} < \frac{1}{n} < 2 \epsilon. $$
So that for any $l \in \{\phi(n), \dots, \psi(n)\}$, one cannot go after $B + \epsilon$ without a jump in the neighborhood $(B-\epsilon, B+\epsilon)$ of $B$.
This shows that for any $n$, it exists $k \in \{\phi(n), \dots, \psi(n) \}$ such that $d_k \in (B - \epsilon, B+\epsilon)$. 
