Let $p:\tilde{X} \mapsto X$ be a covering map with $\tilde{X}$ path connected. Why are all liftings of a closed path $f$ in $X$ either closed or not closed? If $\omega$ is a path from $\tilde{x_0}$ to $\tilde{x_1}$, both in the pre-image of $x_0$ in $X$, $p\omega$ will be a closed path based at $x_0$ in $X$. The path homo topy lifting theorem then guarantees that all liftings of $p\omega$ starting at $\tilde{x_0}$ equal $\omega$. But why can't there be another closed lift based at another point in the pre-image of $x_0$?
Edit: $p$ is a regular covering
In a regular covering, the group of deck transformations acts transitively on the fiber of any chosen point. See Hatcher's Algebraic Topology page 70-71 (he calls regular coverings, normal.)
So, suppose we have a loop $\gamma$ in $X$ with $\gamma(0) = \gamma(1) = x$. Let us take two different lifts $\gamma', \gamma''$ using two different points $x', x''$ in $p^{-1}(x)$. By the first statement, there is a deck transformation $F : \tilde{X} \rightarrow \tilde{X}$ that takes $x'$ to $x''$. Now, we have $$p \circ F \circ \gamma' = p \circ \gamma' = \gamma$$ because $p \circ F = p$ by definition of deck transformation. Hence $F \circ \gamma'$ is a path that lifts $\gamma$ and is based on $x''.$ By uniqueness of lifts of paths, $\gamma'' = F \circ \gamma'.$
Now, it is clear from the above argument that $\gamma''(0) = F(\gamma'(0))$ and $\gamma''(1) = F(\gamma'(1))$. Since $F$ is a homeomorphism and hence a bijection, we see that $\gamma'$ is a loop if and only if $\gamma''$ is a loop.