Hints for this result (for a more specific case) were given here, but I want to give a formal proof. Please verify whether my proof is correct -- any general feedback is also much appreciated!
Claim
Let $G$ be a topological group and $H$ be a discrete normal subgroup. Then $\pi : G \rightarrow G/H $ is a covering map.
Proof
By the discreteness of $H$, there exists an open neighborhood $V$ of $1$ such that $V \cap H = \{1\}$. It follows that $\pi|_U~U \xrightarrow \sim \pi(U)$ is a homeomorphism.
Claim 1: $U \cap hU$ if $h \neq 1$. Define $f : G \times G \rightarrow G$ by $(x,y) \mapsto x y^{-1}$. Since G is a topological group, f is continuous. Because $V$ is open, so is $f^{-1}(V)$. Since $(1,1) \in f^{-1}(V)$, there exists an open neighborhood $W$ of $(1,1)$ such that $W \subset f^{-1}(V)$. Taking into account the product topology, there exists a $U$ open in $G$ such that $(1,1) \in U \times U \subset W \subset f^{-1}(V)$. Thus, $f(U \times U) \subset V$.
Now suppose $x \in U \cap hU$. Then $u_1 = x = hu_2$ for some $u_i \in U$. But then $h = u_1 u_2^{-1} \in V \cap H = {1}$, contradicting the hypothesis that $h \neq 1$. $\square$
Claim 2: $\pi^{-1} (\pi(U)) = HU = \{hU : h \in H \}$.
($\subset$) \begin{align} V \subset \pi^{-1}(\pi(U)) &\Longrightarrow \pi(V) \subset \pi(U) \\ &\Longrightarrow \pi(v) = \pi(u)~~\forall v \in V, u \in U \\ &\Longrightarrow v = hu~~\text{for some}~v \in V, u \in U. \end{align} ($\supset$)
If $hu \in HU$ then $\pi(hu) = \pi(u) \in \pi(U)$. So $hu \in \pi^{-1}(\pi(U))$. $\square$
Note that $gH \subset \pi(gU)$ for each $g \in G$. Since $gU$ is a translate of $U$, it also contains only one element of $H$. Thus, one follows the same steps to prove Claims 1 and 2 and we obtain the desired result. $\square$
