I am reading "A Guide to the Classification Theorem for Compact Surfaces" by Jean Gallier and Dianna Xu. In Appendix E, page 163, I am having trouble understanding the proof of Lemma E.2.
$\textbf{Lemma E.2:}$ Let $\gamma, \gamma_2, \gamma_3 : [0, 1] \rightarrow \mathbb{R}^2$ be three continuous simple closed curves, and suppose $\gamma_3([0, 1]) \subseteq \text{int}\gamma_2([0, 1])$. Define a $\textit{bad segment}$ to be a segment of $\gamma$ which joins 2 points of $\gamma_2([0, 1])$ and has all other points in $\text{int}\gamma_2([0, 1])$. Define a $\textit{very bad segment}$ to be a bad segment which intersects $\gamma_3([0, 1])$. Then there are finitely many very bad segments.
$\textbf{Proof:}$ Suppose for contradiction that infinitely many bad segments intersect $\gamma_3([0, 1])$. Let $P_1, P_2, ...$ be an infinite sequence of very bad segments. Each very bad segment $P_n$ corresponds to two distinct points, $p_n=\gamma(u_n)$ and $q_n =\gamma(v_n)$. Define the sequence $(t_i)$ by $t_{2i}= u_i, t_{2i+1}=v_i$. Let $t$ be an accumulation point of $(t_i)$.
$\textbf{$\gamma$ is continuous, so some subsequences of $(p_n)$ and $(q_n)$ both converge to $s= \gamma(t)$.}$
Note that $\gamma$ and $\gamma_2$ intersect at $s$. Since $\gamma$ is continuous, for any $\eta >0$, there is some $\epsilon >0$ such that if $|t-u|<\epsilon$, then $||s-\gamma(u)||<\eta$. So there is some $P_n \subseteq B_\eta (s)$. Take $\eta = d(\gamma_3([0, 1]), \gamma_2([0, 1]))$ such that there is some $P_n$ which does not intersect $\gamma_3([0, 1])$ -- a contradiction.
$\square$
$\textbf{My Confusion:}$ I highlighted in bold the sentence which is giving me trouble. I do not see why $(p_n)$ and $(q_n)$ must share an accumulation point.
The idea is to think about the length $L_i = |u_i - v_i|$ of the domain interval for $P_i$. These domain intervals are subintervals of $[0,1]$ with disjoint interiors, so the sum of their lengths is a convergent series $\sum_{i=1}^\infty L_i$. It follows that the sequence $L_i=|u_i-v_i|$ converges to zero.
Now pick any subsequence $(u_{i_n})$ of the $(u_i)$ sequence that converges to a limit $t \in [0,1]$; it follows that the corresponding subsequence $(v_{i_n})$ of the $(v_i)$ sequence converges to the same limit $t \in [0,1]$.
Therefore, by continuity of $\gamma$, the two sequences of points $\gamma(u_{i_n})$ and $\gamma(v_{i_n})$ converge to the same limit $\gamma(t)$.