Discrete subgroup implies closed

Statement
Any discrete subgroup of a T0 topological group (i.e., a subgroup that is discrete in the subspace topology), is a closed subgroup.

Facts used

 * Every open subset of a topological group containing the identity has a symmetric open squareroot: If $$U$$ is an open subset of a topological group containing the identity element of the topological group, there exists an open subset $$V$$ such that $$V$$ is symmetric: it contains the identity element and is closed under the inverse map, and further, such that $$V.V \subset U$$

Proof
Given: A T0 topological group $$G$$, a discrete subgroup $$H$$

To prove: $$H$$ is a closed subgroup of $$G$$

Proof: Let $$e$$ denote the identity element of $$G$$. Since $$H$$ is discrete, there exists an open set $$U \ni e$$ such that $$U \cap H = \{ e \}$$. By the fact stated above, there exists a symmetric open subset $$V$$ such that $$V.V \subset U$$.

Now, suppose $$H$$ is not closed. Then there exists an element $$g \in G$$ such that every open subset containing $$g$$ intersects $$H$$. This yields that every open subset containing the identity intersects $$g^{-1}H$$. In particular, $$V$$ intersects $$g^{-1}H$$. Note that since $$V$$ does not intersect $$H$$, $$g \notin H$$. Hence, $$y \ne e$$. Thus, $$V \setminus \{ y \}$$ is also an open subset containing the identity, and hence it again intersects $$g^{-1}H$$. Thus, we can find another point $$z \in V \cap g^{-1}H$$.

Now consider $$y^{-1}z$$. This is an element of $$H$$. Moreover, since $$V$$ is symmetric, $$y^{-1} \in V$$, so $$y^{-1}z \in V.V \subset U$$. Finally, since $$y \ne z$$, $$y^{-1}z \ne e$$. Thus, we have found a non-identity element in $$U \cap H$$, a contradiction.