When a covering map is finite and connected, there exists a loop none of whose lifts is a loop. I've read the following exercise.

Let $p:\tilde X\to X$ be finite connected covering map. Show that there exists a loop in $X$ none of whose lifts is a loop.

I can't understand why it's supposed to be true. Can't I take $\tilde X=X$ and $p=\mathrm{id}_X?$ Then if I take any loop $\omega$ in $X$, I have $\omega\circ p=\omega$, so $\omega$ is its own lift. Why is it incorrect?
(I'm not sure what a connected covering map is, but I think my example must satisfy this condition, whatever it means, if $X$ is connected.)
Added. Here's what I've got after David Speyer's comments. First of all, we should add to the hypothesis that the degree of $p$ is not $1$. Now, there are the following two theorems.


*

*If $p:\tilde X\to X$ is a covering map, $x_0\in X,\ \tilde x_0\in p^{-1}(x_0),$ then the induced homomorphism $p_*:\pi(\tilde X,\tilde x_0)\to\pi(X,x_0)$ of the fundamental groups is a monomorphism.

*If $p:\tilde X\to X$ is a connected covering map, then its degree is equal to the index $[\pi_1(X,x_0):p_*(\pi_1(\tilde X,\tilde x_0))].$


Suppose every loop in $X$ has a lift that is a loop. This means that for each loop $\omega$ in $X$ there exists a loop $\tilde\omega$ in $\tilde X$ such that $\tilde\omega\circ p=\omega.$ But this means that $p_*([\tilde\omega])=[\omega],$ where $[\gamma]$ denotes the homotopy class of a loop $\gamma.$ But then $p_*$ is a group epimorphism. By the first theorem it is also a monomorphism, and so it is an isomorphism between $\pi_1(\tilde X,\tilde x_0)$ and $\pi_1(X,x_0).$ Therefore, $[\pi_1(X,x_0):p_*(\pi_1(\tilde X,\tilde x_0))]=1,$ and, by the second theorem, the degree of $p$ is $1$, a contradiction with the added hypothesis.
Is this correct? I think it's probably not, because I'm not using the finiteness of $p$. The finiteness of $p$ means, according to the theorem that the index mentioned before is finite. But I can't see at all why that should be necessary.
The only example I know of an infinite covering map is $e^{it}$ covering the unit circle with the real line. But here there are plenty of loops that have no "loopy" lifts because there are no nontrivial loops in the real line, and there are many in the circle. So if the infinity of the covering map is indeed a problem, this example doesn't show it.
Re David Speyer's answer
Don't the theorems work for any $\tilde x_0\in p^{-1}(x_0)?$ That is, for each such $\tilde x_0,$ we have a monomorphism $p_{*,\tilde x_0} : \pi_1(\tilde X,\tilde x_0) \to\pi_1(X,x_0).$ If the hypotheses of the problem hold, there is, as you say, a $y\in p^{-1}(x_0)$ and a loop $\tilde\omega$ in $\tilde X$ starting and ending in $y$ such that $p\circ\tilde\omega=\omega$. Then can't I just use $p_{*,y}$ as above?
 A: Your proposed proof doesn't work. Let me write out the key part in more detail, so that you can see the issue.
Recall that an element of $\pi_1(X, x_0)$ is an equivalence class of loops starting from and ending at $x_0$. Let $\tilde{X} \to X$ be a connected covering map, and let $\tilde{x}_0$ in $\tilde{X}$ be a preimage of $x_0$. As you say, $\pi_1(\tilde{X}, \tilde{x}_0) \to \pi_1(X, x_0)$ is injective. You then try to prove that it is surjective, but your proof is flawed as I will now explain.
Let $\omega$ be a loop in $X$, starting and ending at $x_0$. By hypothesis, there is a loop $\widetilde{\omega}$ in $\tilde{X}$ mapping to $\omega$. However, $\widetilde{\omega}$ need not start and end at $\tilde{x}_0$. The start and end of $\widetilde{\omega}$ is some preimage of $x_0$, call it $y$, but we are not assuming that $y=\tilde{x}_0$.
The essence of this problem is (1) to work out what consequence the hypothesis has for the fundamental groups, and (2) do some nontrivial group theory. 
A hint for part (1): Let $\eta$ be a path in $\tilde{X}$ from $\tilde{x}_0$ to $y$. Then the concatenation $\eta \widetilde{\omega} \eta^{-1}$ is a class in $\pi_1(\tilde{X}, \tilde{x}_0)$. What implication does this have for the map of fundamental groups $\pi_1(\tilde{X}, \tilde{x}_0) \to \pi_1(X, x_0)$?
But I continue to be confused as to how this problem could be reasonably assigned in a class where the only infinite covering map you have seen is the circle to the real line.
A: If the covering space is path-connected, then for $\widetilde x_0$ and $\widetilde x_1$ in the fiber of $x_0$ there is a path $\widetilde h$. Now the $p_*(\pi_1(\widetilde X,\widetilde x_1))$ and $p_*(\pi_1(\widetilde X,\widetilde x_0))$ are related in a special way. A loop $\widetilde\gamma$ at $\widetilde x_1$ can be written as $\overline{\widetilde h}\cdot\widetilde\beta\cdot\widetilde h$ where $\widetilde\beta=(\widetilde h\cdot\widetilde\gamma\cdot\overline{\widetilde h})$ is a loop at $\widetilde x_0$. Denote with $\beta,\gamma,h$ the homotopy classes of the images under $p$. Then the image of $[\gamma]$ under $p_*$ is $h^{-1}\beta h$. Analogously, the image of $[\beta]$ is $h\gamma h^{-1}$. So we have $p_*(\pi_1(\widetilde X,\widetilde x_1))=h^{-1}p_*(\pi_1(\widetilde X,\widetilde x_0))h$. You want to show that the lift at every point in the fiber of $x_0$ is not a loop. So you have to prove that there is a loop in $\pi_1(X,x_0)$ which is not in any conjugacy class of $p_*(\pi_1(\widetilde X,\widetilde x_0))$. Since the image has finite index, we can apply Arturo's result that the union of the conjugates is not the whole group. That should do it.
