My proposed proof for Question 3 of section 1.3 of Hatcher:
Let $p : \tilde{X} \to X$ be a covering space with $p^{-1}(x)$ finite and non-empty for all $x \in X$. Show that $\tilde{X}$ is compact Hausdorff if and only if $X$ is compact Hausdorff.
Suppose $X$ is compact Hausdorff. Consider an open cover of $\tilde{X}$ given by $$\tilde{X} = \bigcup_{\alpha} U_{\alpha}.$$ We want to show that such a cover has a finite subcover. Since $X$ is compact, for every open cover of $X$ there is a finite subcover. Therefore, consider that for each $x \in X$, there is an open neighbourhood $U_x$. This enables us to cover $X$ by such neighbourhoods. Therefore, $$X = \bigcup_{x \in X} U_x.$$ By compactness, there are finite number of $x \in X$ such that this cover is finite. We note that $p : \tilde{X} \to X$ is continuous and therefore $p^{-1}(U_x)$ is open in $\tilde{X}$ for each $x$. We assume that $p^{-1}(x)$ is finite, which yields that $p^{-1}(U_x)$ is finite for each $x$. It is therefore possible to write $\tilde{X}$ as the union of a finite number of such $p^{-1}(U_x)$ since $X$ is assumed to be compact.
Now we must show that $\tilde{X}$ is Hausdorff. Let $x_1, x_2 \in X$ be such that $x_1 \neq x_2$. Since we have assumed that $X$ is Hausdorff, it follows that there exist neighbourhoods $U_1, U_2$ of $x_1,x_2$ respectively, such that $U_1 \cap U_2 = \emptyset$. By then considering $p^{-1}(U_1)$ and $p^{-1}(U_2)$ it follows from the continuity of $p$ that $p^{-1}(U_1) \cap p^{-1}(U_2) = \emptyset$ for $p^{-1}(x_1) \neq p^{-1}(x_2)$. We therefore conclude that $\tilde{X}$ is Hausdorff.
Conversely, suppose $\tilde{X}$ is compact Hausdorff. Since $p^{-1}(x)$ is non-empty for each $x \in X$, this yields that $p : \tilde{X} \to X$ is a continuous surjection. $X$ is therefore compact. To show that $X$ is Hausdorff, consider two distinct points $x_1, x_2 \in X$. Then $p^{-1}(x_1)$ and $p^{-1}(x_2)$ are disjoint, non-empty and finite. Since $\tilde{X}$ is Hausdorff there exist open neighbourhoods $U_1$ and $U_2$ of $p^{-1}(x_1)$ and $p^{-1}(x_2)$ respectively such that $U_1 \cap U_2 = \emptyset$. Let $E_1, E_2$ be even coverings of $x_1,x_2$ respectively. Since $p: \tilde{X} \to X$ is a covering space, we have that $p^{-1}(E_1), p^{-1}(E_2)$ are unions of disjoint open subsets such that each open subset is homeomorphic to the even cover. We can therefore write $$p^{-1}(E_1) = \bigcup_{j=1}^n V_j \ \ \text{and} \ \ p^{-1}(E_2) = \bigcup_{k=1}^m W_k.$$ We need to ensure that $$ \Bigg (\bigcap_{j=1}^n p(V_j \cap U_1) \Bigg ) \bigcap \Bigg ( \bigcap_{k=1}^m p(W_k \cap U_2) \Bigg ) = \emptyset.$$ This is easily verified however from the face that $\bigcap_{j=1}^n (V_j \cap U_1) \subset U_1$ and $\bigcap_{k=1}^m (W_k \cap U_2) \subset U_2$ and $U_1 \cap U_2 = \emptyset$.