Let $S=\{t\in T\ :\ \tilde{F}_1(t)=\tilde{F}_2(t)\}$. We need to prove that $S$ is open and that $T\setminus S$ is open (i.e. $S$ is closed). Since $T$ is connected, this implies that either $S=\emptyset$ or $T\setminus S=\emptyset$, so if $S$ is nonempty then $S=T$ and $\tilde{F}_1\equiv\tilde{F}_2$.

(5.22) Given a point $t\in T$ and let $x=F(t)$. Since $p\colon Y\to X$ is a covering space, there is an elementary neighourhood $W\subset X$ of $x$. In $Y$ there are elementary sheets $V_1$ and $V_2$, projecting to $W$, containing $\tilde{F}_1(t)$ and $\tilde{F}_2(t)$ respectively.

(7.32) Since $\tilde{F}_1,\tilde{F}_2$ are continuous, there exist open sets $U_1,U_2\subset T$ such that $t\in U_i$ and $\tilde{F}_1(U_i)\subset V_i$ for $i=1,2$. Let $U=U_1\cap U_2$.

(8.45) We want to show that $T\setminus S$ is open, so suppose that $t\in T\setminus S$. Then I claim that $U\subset T\setminus S$, which will prove that every point in $T\setminus S$ has an open neighbourhood contained in $T\setminus S$, which implies that $T\setminus S$ is open.

(9.40) The claim can be proved as follows. Since $t\in T\setminus S$, $\tilde{F}_1(t)\neq\tilde{F}_2(t)$, so $V_1$ and $V_2$ are disjoint. Therefore $\tilde{F}_1(t')\neq\tilde{F}_2(t')$ for all $t'\in U$ (because $t'\in U$ implies $\tilde{F}_1(t')\in V_1$ and $\tilde{F}_2(t')\in V_2$, and $V_1\cap V_2=\emptyset$).

(11.25) We also want to show that $S$ is open, so suppose that $t\in S$. Then $\tilde{F}_1(t)=\tilde{F}_2(t)$, so $V_1=V_2=:V$. There exists a local inverse $q\colon W\to V$ to $p$, i.e. $p\circ q=id_W$, $q\circ p=id_V$.

(12.53) If $t'\in U$ then $\tilde{F}_1(t')\in V$ and $\tilde{F}_2(t')\in V$, so $q(F(t'))=q(p(\tilde{F}_1(t')))=\tilde{F}_1(t')$ and $q(F(t'))=q(p(\tilde{F}_2(t')))=\tilde{F}_2(t')$, so we deduce that $\tilde{F}_1(t')=\tilde{F}_2(t')$. Therefore $\tilde{F}_1$ and $\tilde{F}_2$ agree on $U$, which proves that $S$ is open.

This proves the lemma.