Partitioning the Domain of a Curve Here's an instance of a very common technique in proofs involving continuous functions out of a closed inverval $[a,b] \in \mathbb{R}$: Let $M$ be a smooth manifold and $\gamma : [a,b] \to M$ a smooth curve. The image of $\gamma$ might not be contained in a single smooth chart on $M$, but I've read that the compactness of $[a,b]$ allows us to find a finite partition $a = a_0 < a_1 < \cdots < a_k = b$ such that $\gamma[a_i,a_{i+1}]$ is contained in a smooth chart for all $i$.
Here's my proof for this: Cover $\gamma[a,b] \in M$ with smooth charts $U_\alpha$, so that the sets $\gamma^{-1}(U_\alpha)$ are an open cover of $[a,b]$. We can then find a Lebesgue number $\delta$ for this cover of $[a,b]$ and select our $a_i$ so that $a_{i+1} - a_i < \delta$ for all $i$. (Correct me if this proof is wrong.) However the book I'm reading does not mention Lebesgue numbers; is there a simpler way to prove this without using them?
 A: The result to be proved is so close to Lebesgue's lemma that it is hard to distinguish between the two results, but one can give a fairly natural proof of the desired result without mentioning Lebesgue's lemma explicitly - if that's of any use.
If $\mathscr{U}$ is an open cover of $[a, b],$ then $[a, b]$ is covered by the set of all open intervals $I$ such that $I \subseteq U$ for some $U \in \mathscr{U}.$ By compactness, therefore, $[a, b]$ is covered by finitely many such intervals.
Let $\mathscr{I}$ be such a finite set of intervals, and let $E$ be the union of $\{a, b\}$ with the set of endpoints (either upper or lower) of intervals $I \in \mathscr{I}$ belonging to the open interval $(a, b).$
Let $p, q$ be two successive elements of the totally ordered finite set $E.$
Pick any point $r$ such that $p < r < q.$
By the definition of $\mathscr{I},$ there exists an open interval $I$ such that $p \in I \subseteq U$ for some $U \in \mathscr{U},$ and there exists an open interval $J$ such that $q \in J \subseteq V$ for some $V \in \mathscr{U}.$
By the definitions of $E$ and $p$ and $q,$ the upper endpoint of $I$ is at least $q > r,$ therefore $[p, r] \subset I \subseteq U \in \mathscr{U}.$
Similarly, because the lower endpoint of $J$ is at most $p < r,$ we have $[r, q] \subset J \subseteq V \in \mathscr{U}.$
Therefore the set $E$ together with a choice of a point $r$ for each pair $(p, q)$ of successive points of $E$ constitutes a partition of $[a, b]$ whose every closed subinterval is contained in a single set belonging to the given cover $\mathscr{U}.$
Applying this result to the covering of $[a, b]$ by the sets $\gamma^{-1}(U_\alpha),$ we find a partition of $[a, b]$ for whose every closed subinterval $K$ there exists $\alpha$ such that $\gamma(K) \subseteq U_\alpha.$ 
