Two admissible continuation chains along one path admit a common refinement

By [L1], each set ${\gamma }^{-1}(U_i)$ is open in $[0,1]$, and the containment $\gamma ([t_i,t_{i+1}])\subseteq U_i$ gives $[t_i,t_{i+1}]\subseteq {\gamma }^{-1}(U_i)$. Hence the finite family $\mathcal{U}:=\{{\gamma }^{-1}(U_i):0\le i<m\}$ is an open cover of $[0,1]$. Likewise $\mathcal{V}:=\{{\gamma }^{-1}(V_j):0\le j<n\}$ is an open cover of $[0,1]$.

Apply [L2] to the compact interval $[0,1]$ and the two open covers $\mathcal{U}$ and $\mathcal{V}$. Let ${\delta }_{\mathcal{U}},{\delta }_{\mathcal{V}}>0$ be corresponding Lebesgue numbers, put $\delta :=\min \{{\delta }_{\mathcal{U}},{\delta }_{\mathcal{V}}\}$, and choose a subdivision $0=u_0<u_1<\cdots <u_r=1$ whose mesh is less than $\delta$. Then every interval $[u_k,u_{k+1}]$ has diameter less than both ${\delta }_{\mathcal{U}}$ and ${\delta }_{\mathcal{V}}$.

For each $k<r$, the interval $[u_k,u_{k+1}]$ has diameter less than ${\delta }_{\mathcal{U}}$ and less than ${\delta }_{\mathcal{V}}$, so the Lebesgue-number property gives indices $i,j$ with $[u_k,u_{k+1}]\subseteq {\gamma }^{-1}(U_i)$ and $[u_k,u_{k+1}]\subseteq {\gamma }^{-1}(V_j)$. Equivalently, $\gamma ([u_k,u_{k+1}])\subseteq U_i\cap V_j$. Restricting $f_i$ and $g_j$ to these smaller intervals produces the required common refinement.