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How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in $\mathbb{R}$ is non-empty. Here goes the proof:
We construct recursively a nested sequence $I_j := [a_j, b_j]$ of closed intervals for $j \geq 0$. Let $I_0 := [0,1]$. For every $j \geq 0$, construct $I_{j+1}$ as follows: let $m_j$ be the midpoint of $I_j$. If the curves intersect at $t = m_j$, then we are done, so stop the sequence. Otherwise set $I_{j+1}$ to be $[a_j, m_j]$ or $[m_j, b_j]$ depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that $c_1$ is to the "left" of $c_2$ at $t = a_j$ and to the "right" of $c_2$ at $t = b_j$).
If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that $\cap_{j=0}^\infty I_j = {t}$ \lbrace t\rbrace$for some$t \in [0,1]$. Then we must have$c_1(t) = c_2(t)$. 1 How about the following, using the Nested Intervals Theorem (which was in my 2nd year Calculus text) which says the intersection of a nested sequence of closed intervals in$\mathbb{R}$is non-empty. Here goes the proof: We construct recursively a nested sequence$I_j := [a_j, b_j]$of closed intervals for$j \geq 0$. Let$I_0 := [0,1]$. For every$j \geq 0$, construct$I_{j+1}$as follows: let$m_j$be the midpoint of$I_j$. If the curves intersect at$t = m_j$, then we are done, so stop the sequence. Otherwise set$I_{j+1}$to be$[a_j, m_j]$or$[m_j, b_j]$depending on whether the curves "switch from left to right" on the first sub-interval or the 2nd (let's say you always make sure that$c_1$is to the "left" of$c_2$at$t = a_j$and to the "right" of$c_2$at$t = b_j$). If the sequence is finite, then the curves must intersect, as noted above. So assume the sequence is infinite. The Nested Intervals Theorem and the fact that the length decreases by a factor of 2 at every step implies that$\cap_{j=0}^\infty I_j = {t}$for some$t \in [0,1]$. Then we must have$c_1(t) = c_2(t)\$.