Let $M$ be a manifold, $\alpha:[0, 1]\rightarrow M$ a continuous curve and $\Gamma$ an open cover of $Im \alpha$ in $M$.
Can we extract a finite subset of $\Gamma$ which is an open cover of $Im \alpha$ such that if ${U}_{i},{U}_{i+1}, {U}_{i+2}\in \Gamma$ then ${U}_{i}\cap {U}_{i+1}\neq \emptyset$ and ${U}_{i+1}\cap{U}_{i+2}\neq \emptyset$, that is, the ${U}_{i}$ are chained? If so, how is this done?
Here is a standard way to do this by applying the Lebesgue number lemma, a common argument in topology.
Consider $\Gamma' =\{\alpha^{-1}(U) \,|\, U \in \Gamma\}$ which is an open cover of $[0,1]$.
Since $[0,1]$ is compact, we may apply the Lebesgue number lemma, which gives us a number $\lambda>0$ such that for any subset $A \subset [0,1]$, if the diameter of $A$ is no larger than $\lambda$ then there exists $\alpha^{-1}(U) \in \Gamma'$ such that $A \subset \alpha^{-1}(U)$; equivalently, there exists $U \in \Gamma$ such that $\alpha(A) \subset U$.
Now choose an integer $n > \frac{1}{\lambda}$ and subdivide the interval $[0,1]$ into $n$ equal portions each of length $1/n$: $$[0,1/n] \cup [1/n,2/n] \cup [2/n,3/n] \cup \cdots \cup [(n-1)/n,1] $$ Since $[(k-1)/n,k/n]$ has length $\frac{1}{n} < \lambda$, there exists $U_k \in \Gamma$ such that $\alpha[(k-1)/n,k/n] \subset U_k$.
The sequence $U_1,...,U_n$ covers the image of $\alpha$, and $U_{k-1} \cap U_k$ is nonempty because it contains $\alpha(k/n)$.
