If fundamental group of a compact path connected space is finite then its universal cover is compact Let $X$ be a compact and path connected space and let $p:\tilde X\rightarrow X$ be its universal cover. I can show that if $\tilde X$ is compact then $\pi_{1}(X)$ is finite:
$\pi_{1}(X, x_{0})$ acts on $p^{-1}(x_{0})$  via $[\gamma].y=\tilde\gamma(1)$ where $\tilde\gamma$ is the unique lifting strating at y. Another observation is that 
Stab(y)={$[\gamma]$$\in\pi_{1}(X, x_{0})$ : $\tilde\gamma$ is a closed loop}=$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$ where $p_{*}$ is the induced homomorphism of $p$. Since this action is transitive,  there is a bijection between  $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$/$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$. In our case $\tilde X$ is simply connected. So $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$ has the same cardinality. But $p^{-1}(x_{0})$ is closed and discrete subspace of compact space $\tilde X$. so it is finite, as desired.
But I couldn't show the converse. I am new in algebraic topology and I follow Hatcher's book chapter 1. Can any one give a hint or answer? Thanks!
 A: I find the following a useful way of thinking about it.
We may as well prove the statement that a covering map with finite fibers is a proper map (this means that the pre-image of any compact set is compact). (This argument is nice because it easily generalizes to prove the statement that a locally trivial fiber bundle with compact fibers is proper).
First note that when the covering map is just the projection $U \times F \rightarrow U$ then this map is proper because the preimage of a compact set $K \subset U$ is just $K \times F$ which is compact (being the product of compact spaces). In general, suppose that $\widetilde{X} \rightarrow X$ is a cover with finite fiber, $F$, then we may take a cover $\{U_\alpha\}$ of $X$ so that the restriction of the map to the cover just looks like $U_\alpha \times F \rightarrow U_\alpha$, and we've already seen that this is a proper map. 
It is an easy exercise to see that properness (as I've defined it above) is local on the target, i.e. that the map $\widetilde{X} \rightarrow X$ is proper if and only if there is a cover of $X$ where all the restrictions are proper. (Indeed, any compact subset of $X$ is covered by finitely many $U_\alpha$, and so the preimage of $X$ is the union of the preimages of those specific $X \cap U_\alpha$. A finite union of compact sets is compact, QED.)
It follows that $\widetilde{X} \rightarrow X$ is a proper map. But now, if $X$ is compact, then the preimage of $X$, which is $\widetilde{X}$, must be compact, which was to be shown.
A: Let $\{ \tilde{U}_i : i \in I \}$ be an open cover of $\tilde{X}$. For each point $\tilde{x}$ in the fibre $p^{-1} \{ x \}$, choose an open neighbourhood $\tilde{U}_{\tilde{x}}$ of $\tilde{x}$. Since $p$ is a local homeomorphism, we may choose an open neighbourhood $\tilde{V}_{\tilde{x}} \subseteq \tilde{U}_{\tilde{x}}$ of each $\tilde{x}$ that $p$ maps homeomorphically onto an open neighbourhood $V_{\tilde{x}}$ of $x$, and then $V_x = \bigcap_{\tilde{x} \in p^{-1} \{ x \}} V_{\tilde{x}}$ is an open neighbourhood of $x$ such that $p^{-1} V_x$ is a disjoint union of homeomorphic copies of $V_x$, each contained in some $\tilde{U}_i$.
Observe that $p$ is surjective, so $\{ V_x : x \in X \}$ is an open cover of $X$, so has a finite subcover. Consider $\{ p^{-1} V_x : x \in X \}$. This is an open cover of $\tilde{X}$, but not necessarily subordinate to $\{ \tilde{U}_i \}$. Nonetheless, by construction, there is a finite refinement of $\{ p^{-1} V_x : x \in X \}$ that is subordinate to $\{ \tilde{U}_i \}$, and this implies the existence of the desired finite subcover. I leave the details to you.
