Proof:

Let ${\cal O}$ be the open cover of $X$ by evenly covered open sets and $\gamma^{-1}(\cal O)$ be the family

$$\gamma^{-1}({\cal O}) = \{\gamma^{-1}(G)/ G\in {\cal O}\}$$

of open sets covering $[0, 1]$. There is a Lebesgue number $\eta$ for this cover and we choose $n$ to be a natural number such that $1/n < \eta$. Consider the partition

$$\Big\{0, \frac{1}{n}, \frac{2}{n},\dots, \frac{n-1}{n}\Big\}.$$

For each $j = 1, 2, \dots, n$, the piece $\gamma([\frac{j-1}{n}, \frac{j}{n}])$ lies in an evenly covered open set in $X$. In particular if $\gamma_0$ denotes the restriction of $\gamma$ to $[0, 1/n]$ then the image of $\gamma_0$ lies in an open set $G_0 \in {\cal O}.$ The conditions (16.1)-(16.2) say that there is a sheet ${\tilde G}_0$ lying over $G_0$ and containing the point ${\tilde x}_0$. Let $p_0$ denote the restriction of $p$ to the sheet ${\tilde G}_0$ and $q_0^{-1}$ be its inverse. On the sub-interval $[0, 1/n]$, we define

$${\tilde \gamma}_0 = q_0\circ \gamma_0$$

thereby obtaining an initial piece of the desired lift ${\tilde\gamma}$. We shall construct the lift ${\tilde\gamma}$ piece by piece defining it on each subinterval of the partition of $[0, 1]$. In what follows $\gamma_j$ denotes the restriction of $\gamma$ to the sub-interval $[\frac{j}{n}, \frac{j+1}{n}]$. Assume inductively that

$${\tilde\gamma}_j:[\frac{j}{n}, \frac{j+1}{n}] \longrightarrow {\tilde X}$$

has been defined such that

$$p \circ {\tilde\gamma}_j = \gamma_j$$

$${\tilde \gamma}_j(j/n) = {\tilde \gamma}_{j-1}(j/n), j \geq 1.$$

$${\tilde \gamma}_0(0) = {\tilde x}_0$$

For the inductive step we set up the notations for the endpoints of the lift ${\tilde \gamma}_j$ namely, let

$$\gamma_j\Big(\frac{j+1}{n}\Big) = x_{j+1},\quad {\tilde \gamma}_j... ...(\frac{j+1}{n}\Big) = {\tilde x}_{j+1}, \quad p({\tilde x}_{j+1}) = x_{j+1}.$$

Let $G_{j+1}\in {\cal O}$ be an evenly covered neighborhood containing $x_{j+1}$ such that $\gamma$ maps $[\frac{j+1}{n}, \frac{j+2}{n}]$ into $G_{j+1}$ and ${\tilde G}_{j+1}$ be the sheet lying over $G_{j+1}$ containing the point ${\tilde x}_{j+1}$. The restriction of $p$ to ${\tilde G}_{j+1}$ is a homeomorphism with inverse $q_{j+1}$ say, so that $q_{j+1}(x_{j+1}) = {\tilde x}_{j+1}$. We set

$${\tilde \gamma}_{j+1} = q_{j+1}\circ \gamma_{j+1}$$

Then ${\tilde \gamma}_{j+1}$ is continuous, $p\circ {\tilde \gamma}_{j+1} = \gamma_{j+1}$ and

$${\tilde \gamma}_{j+1}\Big(\frac{j+1}{n}\Big) = q_{j+1}(x_{j+1}) = {\tilde x}_{j+1} = {\tilde \gamma}_j\Big(\frac{j+1}{n}\Big)$$

By gluing lemma, the pieces ${\tilde \gamma}_j$ may be glued together to yield a continuous function ${\tilde \gamma}: [0, 1] \longrightarrow {\tilde X}$ such that

$$p\circ {\tilde \gamma} = \gamma, \quad {\tilde \gamma}(0) = {\tilde x}_0.$$

The proof is complete. The uniqueness has been already proved in general.