Let $G$ be a connected matrix Lie group and let $A \in G$. Let $A(t)$ be a continuous path in $G$ such that $A(0) = I$ and $A(t) = A$. Let $V$ be a neighborhood of $I$ in $G$. I am trying to show that because $[0, 1]$ is compact there exists a sequence $\{t_k\}_{k = 0}^m$ such that $t_0 = 0$, $t_m = 1$, $t_k < t_{k + 1}$, and $A(t_{k - 1})^{-1} A(t_k) \in V$.
The problem I am having is that you can have $A \notin V$. Why is $A(t_{k - 1})^{-1} A(t_k) \in V$? Intuitively it makes sense. You can pick each $t_k$ from a separate set from the finite subcover of $[0, 1]$. Make sure that $t_{k + 1} - t_k$ is small enough such that $A(t_{k - 1})^{-1} A(t_k) \in V$, since if $t_{k + 1} - t_{k}$ is small $A(t_{k - 1})^{-1} A(t_k) \approx I$. The finite subcover will need to contain sufficiently small intervals. But how do I prove it more rigorously?
Let $B_{\varepsilon}$ be an open ball around $I$ such that $B_{\varepsilon} \subseteq V$. \begin{align} \lim_{\Delta \to 0} A(t)^{-1} A(t + \Delta) &= I, \end{align} since $A$ is continuous. $A$ is uniformly continuous, since it is defined on a closed bounded set. Hence there exists $\delta > 0$ such that $||A(t)^{-1} A(t + \Delta) - I|| < \varepsilon$ for all $|\Delta| < \delta$. Thus $A(t)^{-1} A(t + c) \in B_\varepsilon$ for all $c \in [0, \delta)$. Let $C_k = [k \delta / 2, (k + 1) \delta / 2)$. $1 \in C_m$ for some $m \in \mathbb{N}$ by the Archimedean property. Let $t_0 = 0$, let $t_k \in C_k$ for $k \in [1, m - 1]$, and let $t_m = 1$. Then $t_k - t_{k - 1} < \delta$. Thus $A(t_{k - 1})^{-1} A(t_{k}) \in V$.