I've stumbled on the following lemma (see 1. below) without proof on a book and I'm having a bit of trouble proving it. Any help would be appreciated.
Let $X$ be an arcwise connected compact metric topological space.
Let $C: [0,1] \rightarrow X$ be a simple arc or a simple closed curve. A segment of $C$ is either $C([a,b])$ or $C([0,a]) \cup C([b,1])$ for a closed subinterval $[a,b]$ of $[0,1]$. For a subset $Y$ of $X$, a $Y$-segment is a a segment of $C$ having only its ends on $Y$.
If $Y_1$ and $Y_2$ are disjoint closed subsets of $X$ and $C$ is a simple arc or a simple closed curve, then there exists only finitely many ($Y_1$ U $Y_2$)-segments of $C$ with one end in $Y_1$ and the other in $Y_2$.
Say that $X=[-2,2]$, $C = [0,1]$ and $Y$ is the usual Cantor Set on $[0,1]$. How many $Y$-segments are there?
EDIT: The answer to number 2 should be countably many, see Brian's comment below.
For $0\le a<b\le 1$ let $S(a,b)=C\big[[a,b]\big]$ and $S'(a,b)=C\big[[0,a]\big]\cup\big[[b,1]\big]$. Let $Y_1$ and $Y_2$ be disjoint closed subsets of $X$, and let $\mathscr{S}$ be the set of $(Y_1\cup Y_2)$-segments having one end in $Y_1$ and the other in $Y_2$.
Suppose that there are infinitely many distinct segments of the form $S(a,b)$ in $\mathscr{S}$. Without loss of generality we may assume that there is a sequence $\langle S(a_n,b_n):n\in\Bbb N\rangle$ in $\mathscr{S}$ such that $C(a_n)\in Y_1$ for each $n\in\Bbb N$. By passing to a subsequence if necessary we may assume that $\alpha=\langle a_n:n\in\Bbb N\rangle$ converges monotonically to some $a\in[0,1]$; note that $C(a)\in Y_1$, since $Y_1$ is closed. If $\alpha$ is increasing, we must then have $a_n<b_n<a_{n+1}$ for each $n\in\Bbb N$: clearly $b_n\ne a_{n+1}$, since $C(b_n)\in Y_2$ and $C(a_{n+1})\in Y_1$, and $b_n\not>a_{n+1}$, since $S(a_n,b_n)\in\mathscr{S}$ and $C(a_{n+1})\in Y_1$. Similarly, if $\alpha$ is decreasing, we must have $a_{n+1}<b_{n+1}<a_n$ for each $n\in\Bbb N$. In either case $\langle b_n:n\in\Bbb N\rangle\to a$, and hence $\langle C(b_n):n\in\Bbb N\rangle\to C(a)\in Y_1$, which is impossible: $C(b_n)\in Y_2$ for each $n\in\Bbb N$, and $Y_2$ is closed, so $C(a)\in Y_1\cap Y_2=\varnothing$. Thus, $\mathscr{S}$ contains at most finitely many segments of the form $S(a,b)$.
Now suppose that there are infinitely many distinct segments of the form $S'(a,b)$ in $\mathscr{S}$. As before we may assume that there is a sequence $\langle S'(a_n,b_n):n\in\Bbb N\rangle$ in $\mathscr{S}$ such that $C(a_n)\in Y_1$ for each $n\in\Bbb N$, and $\alpha=\langle a_n:n\in\Bbb N\rangle$ converges monotonically to some $a\in[0,1]$. If $a_n=a_{n+1}$ for some $n\in\Bbb N$, then either $b_n<b_{n+1}$ in which case $C(b_{n+1})$ is a non-endpoint of $S'(a_n,b_n)$ in $Y_2$, or $b_{n+1}<b_n$, in which case $C(b_n)$ is a non-endpoint of $S'(a_{n+1},b_{n+1})$ in $Y_2$. This is impossible, so $\alpha$ must be strictly monotonic. If $\alpha$ is decreasing, then $C(a_{n+1})$ is a non-endpoint of $S'(a_n,b_n)$ in $Y_1$, and if $\alpha$ is increasing, then $0\le a_0<a_1<a_2$, so $C(a_1)$ is a non-endpoint of $S'(a_2,b_2)$ in $Y_1$ even if $b_2=1$ and $C(0)=C(1)$. Again this is impossible, so $\mathscr{S}$ must contain only finitely many segments of the form $S'(a,b)$ and hence only finitely many distinct segments altogether.