Closure of Subgroup is Group

Theorem
Let $G$ be a topological group.

Let $H\leq G$ be a subgroup.

Let $\overline H$ denote its closure.

Then $\overline H$ is a subgroup of $G$.

Proof
We use the One-Step Subgroup Test.

Because $H \subset \overline H$, $\overline H$ is non-empty.

Let $a, b \in \overline H$.

Let $U$ be a neighborhood of $a b^{-1}$.

Let the mapping $f: G\times G \to G$ be defined as:
 * $f\left({x, y}\right) = x y^{-1}$

By definition of topological group, $f$ is continuous.

By definition of product space, there exist neighborhoods $A, B$ of $a, b$ respectively such that:
 * $A \times B \subset f^{-1} \left({U}\right)$

By assumption, there exist $x \in A \cap H$ and $y \in B \cap H$.

Then:
 * $x y^{-1} \subset U \cap H$

Because $U$ was arbitrary:
 * $a b^{-1} \in \overline H$

By One-Step Subgroup Test, $\overline H$ is a subgroup of $G$.