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?
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 $\gamma(K) \subseteq U_\alpha.$