I don't understand the hint (in bold) to prove this problem.
Problem:
Let $f: D \rightarrow \mathbb{R}$ and let $c$ be an accumulation point of D. Then the following are equivalent:
(a) $f$ does not have a limit at $c$.
(b) There exists a sequence $(s_n)$ in $D$ with each $s_n \ne c$ such that $(s_n)$ converges to $c$, but $f((s_n))$ is not convergent in $\mathbb{R}$.
Hint:
To prove that $(a) \Rightarrow (b)$, suppose that (b) is false. Let $s_n$ be a sequence with $s_n \rightarrow c$. Before we can use Theorem 20.8, we must show that given any sequence $(t_n)$ in $D$ with $t_n \Rightarrow c$, we have limit $f(t_n)=L$.
We only know from the negation of (b) that $(f(t_n))$ is convergent. To see that $\lim f(t_n) = L$, consider the sequence $(u_n) = (s_1, t_1, s_2, t_2...)$ and note that $(f(s_n))$ and $(f(t_n))$ are both subsequences of $(f(u_n))$.
My question:
The hint wants for any sequence $(t_n)$. But, how does $(u_n)$ help showing $(t_n)$ is any sequence? What do I need $(u_n)$ for?
The proof is here under Theorem 2.
You have some fixed sequence $(s_n)$ such that $s_n\to c$. Then take any other sequence $(t_n)$ such that $t_n\to c$. By negating $(b)$ we know that $(f(s_n))$ and $(f(t_n))$ are both convergent, but the critical part is to notice that you don't know a priori that the limits are the same. If you knew that the limits are the same, you would prove that $f$ has a limit at $c$.
To prove that $(f(s_n))$ and $(f(t_n))$ converge to the same limit, you consider $(u_n)$ defined as in the text. Obviously, $u_n\to c$. What you now know is that $(f(u_n))$, $(f(s_n))$ and $(f(t_n))$ are all convergent and that $(f(s_n))$ and $(f(t_n))$ are subsequences of $(f(u_n))$. Finally, any subsequence of a convergent sequence is convergent with the same limit. So,
$$\lim_{n\to\infty}f(t_n)=\lim_{n\to\infty}f(u_n)=\lim_{n\to\infty}f(s_n).$$
Since $(t_n)$ was arbitrary, it means that for all sequences $t_n\to c$, $f(t_n)\to L$, where $L = \lim_nf(s_n)$. Therefore, $f$ has a limit at $c$.