Consider the sequence $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$ Consider the sequence 
$$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$$ For which numbers $b$ is there a subsequence converging to $b$?

Your sequence includes every rational number in $(0,1)$ infinitely many times. For any $b\in[0,1]$, there exists a sequence of rational numbers in $(0,1)$ converging to $b$. That sequence necessarily occurs as a subsequence of your sequence. Conversely, any $b\notin[0,1]$ cannot have a subsequence of your sequence converge to it, as there exists a positive number $\epsilon$ such that every element of your sequence is more than $\epsilon$ away from $b$.
Thus, the numbers $b$ for which there is a subsequence converging to $b$ are precisely those $b\in[0,1]$.
