Here $M$ is a topological manifold and $p$ is a fixed point on $M$. And $B$s are basis elements of $M$. I am curious how to construct such finite numbers from $a_0$ to $a_k$. I think I need to use induction but cannot find a way. Could anyone please help me?
Lets start with the open covering $\mathcal{U}=\{U_1,\ldots, U_k\}$ of $[0,1]$. What I'm going to do is I'm going to refine $\mathcal{U}$ in such a way that consecutive elements in $\mathcal{U}$ intersect nonempty and every other intersection is empty.
Refinement 1: since each open subset of $[0,1]$ is a disjoint union of open intervals then (due to compactness) we can assume that each $U_i$ is an open interval. Here by interval I understand a convex subset of $[0,1]$ and thus for example $[0,1/2)$ is an open interval.
Refinement 2: if $U\in\mathcal{U}$ is such that $U\subseteq\bigcup_{V\in\mathcal{U}, V\neq U}V$ then we can remove it from the covering.
With that we obtain that for any $U\in\mathcal{U}$ there is $x\in [0,1]$ such that $x$ belongs only to one element from $\mathcal{U}$: the $U$. I will call such $x$ a $U$-member.
And with that property we can re-order $U_i$'s: $i<j$ if and only if every element from $U_i$ is smaller then a $U_j$-member. This is a total ordering with the following property:
$$U_i\cap U_{i+1}\neq\emptyset$$ $$U_i\cap U_j=\emptyset\text{ for }j>i+1$$
Now it is simple to define the sequence:
$$a_0=0$$ $$a_k=1$$ $$a_j=\text{ any element of }U_{j}\cap U_{j+1}$$
With that obviously $[a_{j-1}, a_j]\subseteq U_j$ and due to ordering of $U_i$'s we have $a_{j-1}<a_j$.
For each open subset $U\subseteq\Bbb [0,1]$, the function $\beta_U:[0,1]\to [0,1]$ defined by $$\beta_U(x)= \begin{cases} 0&x\notin U\\ \sup\{a\geq 0:[x,x+a]\subseteq U\}&x\in U \end{cases}$$ is lower semi-continuous.
Consequently, if $\mathcal U$ is an open cover of $[0,1]$, $\beta=\sup_{U\in\mathcal U}\beta_U$ is lower semi-continuous as well (proof). Thus $\beta$ is a positive lower semi-continuous function over a compact space $[0,1]$, hence it has minimum $m=\min_{[0,1]} \beta>0$ (proof).
Choose a positive integer $k$ such that $\frac 1k<m$, and define $a_i=i/k$ for $0\leq i\leq k$.
