Proof:

We shall construct injective maps from $p^{-1}(x_1)$ into $p^{-1}(x_2)$ and vice versa. Fix a path $\gamma$ in $X$ joining $x_1$ and $x_2$. Pick ${\tilde x}_1 \in p^{-1}(x_1)$ and let ${\tilde\gamma}$ be the lift of $\gamma$ starting at ${\tilde x}_1$ and define a map $T: p^{-1}(x_1)\longrightarrow p^{-1}(x_2)$ by the prescription

$$T: {\tilde x}_1 \mapsto {\tilde \gamma}(1).$$

Likewise let $S: p^{-1}(x_2)\longrightarrow p^{-1}(x_1)$ be the map in the reverse direction constructed using the path $\gamma^{-1}$. Since the inverse path ${\tilde \gamma^{-1}}$ is the unique lift of $\gamma^{-1}$ starting at ${\tilde \gamma}(1)$, we see that

$$S({\tilde \gamma}(1)) = {\tilde \gamma}(0) = {\tilde x}_1,$$

whereby we conclude $S\circ T$ is the identity map on $p^{-1}(x_1)$. By symmetry $T\circ S$ is the identity map on $p^{-1}(x_2)$ as desired. $\square$