Proof:

Since $H$ is discrete, the identity element is not a limit point of $H$ and so there is a neighborhood $U$ of the identity such that $U\cap H = \{1\}$. We may assume $U$ has the property that if $u_1, u_2$ are in $U$ then the product $u_1^{-1}u_2$ is in $U$. This follows from the continuity of the group operation and a detailed verification is left as an exercise. It is easy to see that if $h_1$ and $h_2$ are two distinct elements of $H$ then

$$Uh_1 \cap Uh_2 = \emptyset.$$

Fix $h \in H$ and consider now the set $K$ given by

$$K = \{g \in G\;/\; gh = hg\}$$

We shall show that the subgroup $K$ contains a neighborhood of the identity. Pick a neighborhood $V$ of the identity such that $V = V^{-1}$ and $(hVh^{-1}V)\cap H = \{1\}$. Then for any $g \in V$, we have on the one hand

$$hgh^{-1}g^{-1}\in hVh^{-1}V$$

and on the other hand $hgh^{-1}g^{-1}\in H$ since $H$ is normal. Hence $hgh^{-1}g^{-1} \in (hVh^{-1}V)\cap H = \{1\}$ which shows that $g$ belongs to $K$ and $K$ contains a neighborhood of the unit element. We may now invoke the previous theorem. $\square$