Proof:

Let $G$ be the subset given by $G = \{t\in T/ f_1(t) = f_2(t)\}.$ The set $G$ is non-empty since $t_0 \in G$. We shall show that $G$ is both open and closed in $T$ from which the result would follow since $T$ is connected. For $t \in G$ pick an evenly covered neighborhood $U$ of

$$x = p(f_1(t)) = p(f_2(t)).$$

and ${\tilde U}$ be the sheet lying over $U$ and containing $f_1(t) = f_2(t)$. The set

$$N = f_1^{-1}({\tilde U})\cap f_2^{-1}({\tilde U})$$

is open and contains $t$. If $z \in N$ then $f_1(z)$ and $f_2(z)$ both belong to $\tilde{U}$ and $p(f_1(z)) = p(f_2(z)) = f(z)$. But $p$ restricted to ${\tilde U}$ is injective and so $f_1(z) = f_2(z)$ for all $z \in N$ and we conclude that $N \subset G$. The proof that $G$ is closed is left as an exercise. The student may assume that the spaces involved are Hausdorff (see exercise 7 of lecture 2).