Proof:

We begin with the observation that the condition spelled out in (i) is independent of the choice of $x_0$ and also independent of the choice of ${\tilde x}_0\in p^{-1}(x_0)$. Well, changing the element ${\tilde x}_0$ in the fiber would give a conjugate subgroup but the normality hypothesis says that the conjugacy class of the subgroup $p_*(\pi_1({\tilde X}, {\tilde x}_0))$ is a singleton. Second, a group isomorphism must take a normal subgroup to a normal subgroup and so the condition (i) does not depend on the choice of the base point $x_0$.

Proof that (i) implies (ii). Suppose that $p_*(\pi_1({\tilde X}, {\tilde x}_0))$ is a normal subgroup of $\pi_1(X, x_0)$ and $\gamma$ is an arbitrary loop in $X$ based at $x_0$. Let ${\tilde \gamma}_1$ and ${\tilde \gamma}_2$ be two lifts of $\gamma$ with initial points ${\tilde x}_1, {\tilde x}_2$ such that ${\tilde \gamma}_1$ a closed loop. Then

$${\tilde x}_1\cdot[\gamma] = {\tilde x}_1$$

which means $[\gamma]$ is in the stabilizer of ${\tilde x}_1$ and hence, by (iii) of theorem (17.2), $[\gamma]$ belongs to the stabilizer of ${\tilde x}_2$. Thus

$${\tilde x}_2\cdot [\gamma] = {\tilde \gamma}_2(1) = {\tilde x}_2$$

and we see that ${\tilde \gamma}_2$ is also closed.

Proof that (ii) implies (i). If $p_*(\pi_1({\tilde X}, {\tilde x}_0))$ is not a normal subgroup of $\pi_1(X, x_0)$ then it has at least two distinct conjugates which, by virtue of theorem (17.2), must be the stabilizers of say ${\tilde x}_1, {\tilde x}_2 \in p^{-1}(x_0)$. Thus there exists $[\gamma] \in p_*(\pi_1({\tilde X}, {\tilde x}_1)) = \;$Stab $\;{\tilde x}_1$ but $[\gamma] \notin p_*(\pi_1({\tilde X}, {\tilde x}_2)) = \;$Stab $\;{\tilde x}_2$. In other words

$${\tilde x}_1\cdot[\gamma] = {\tilde \gamma}_1(1) = {\tilde x}_1,\quad {\tilde x}_2\cdot [\gamma] = {\tilde \gamma}_2(1) \neq {\tilde x}_2,$$

where ${\tilde \gamma}_1$ and ${\tilde \gamma}_2$ are the lifts of $\gamma$ starting at ${\tilde x}_1$ and ${\tilde x}_2$ respectively. Thus the lift ${\tilde \gamma}_1$ of $\gamma$ is closed whereas the lift ${\tilde \gamma}_2$ is not closed.