I am attempting to understand a proof that an isotopy of two Legendrian knots $L_0$ and $L_1$ in a closed contact manifold (M,$\xi$) can be extended to an contact isotopy $\phi$ of M such that $\phi_0 = id$ and $\phi_1(L_0) = L_1$. There are two parts in this proof which are unclear to me and I would be grateful for any clarification.
The proof is given in Etnyre's Legendrian and Transversal Knots. It is listed as Theorem 2.12.
We begin by letting $L_t$ $(0 \leq t \leq 1)$ be an isotopy through Legendrian knots in (M,$\xi$). Then there is a family of diffeomorphisms $\phi_t : M \to M$ such that $\phi_t(L_0) = L_t$. He claims we can easily arrange that $\phi_t^*(\xi|_{L_t}) = \xi|_{L_0}$. (I am not sure why this is true? Perhaps we can perturb $\phi_t$ in a tubular neighborhood of $L_0$ so that the normal direction is preserved, but I am not sure this is possible?)
Now set $\xi_t = \phi_t^*(\xi)$. This is a 1-parameter family of contact structures on M and $\xi_t = \xi_0$ along $L_0$ by the condition imposed on $\phi_t$ above. By Gray's Theorem we can find a family of diffeomorphisms $\psi_t$ such that $\psi_t^*(\xi_t) = \xi$ and $\psi_t(L_0) = L_0$. (I am also unclear here as to why $\psi_t(L_0) = L_0$. It should follow from the fact that $\xi_t = \xi$ along $L_0$ but I am not sure how.)
The rest of the proof is straightforward and clear assuming the above.
Thanks for your help.