Theorem: Suppose that $p : X \to Y$ is a covering map.
If $\gamma:[0,1] \to Y$ is continuous, $x_0 \in X$ and $p(x_0) = \gamma(0)$ then there exists a unique continuous function $\tilde{\gamma} : [0,1] \to X$ such that $\tilde{\gamma}(0) = x_0$ and $p \circ \tilde{\gamma} = \gamma$.
To prove existence, let $A$ be the set of $t_0 \in [0,1]$ such that there exists a continuous $\tilde{\gamma} : [0, t_0] \to X$ with $\tilde{\gamma}(0) = x_0$ and $p(\tilde{\gamma}(t)) = \gamma(t)$ for all $t \in [0, t_0]$.
I need to show that $A$ is both closed and open to prove existence, but I don't know how...