Intersection of all neighborhoods of zero is a subgroup

Let $G$ be a topological abelian group. Let $H$ be the intersection of all neighborhoods of zero.

How is $H = \mathrm{cl}(\{0\})$? Isn't the closure of a set $A$ the smallest closed set containing $A$ which is the same as the intersection of all closed sets containing $A$? But neighborhoods in $G$ are not necessarily closed. Thanks.

(Edit)

To see that $H$ is a subgroup:

First note that by construction $H$ contains $0$.

Furthermore, $f: x \mapsto -x$ is continuous and its own inverse so that $f$ is also open. Hence $U$ is a neighborhood of $0$ if and only if $-U$ is. Now let $x \in H$. Then $x$ is in every neighborhood $U$ of $0$. Hence $x$ is in every neighborhood $-U$ of $0$. Hence $-x$ is in $U$ and hence in $H$.

Alternatively one can verify it as follows: $$x \in H \iff x \in \bigcap_{U \text{ nbhd of } 0} U \iff x \in \bigcap_{U \text{ nbhd of } 0} -U \iff -x \in \bigcap_{U \text{ nbhd of } 0} U \iff -x \in H$$

To see that $x+y$ is in $H$ if $x,y \in H$, note that $g: (x,y) \mapsto x+y$ is continuous. Now let $V$ be an arbitrary neighborhood of $0$. Then since $g$ is continuous there exists a neighborhood $N \times M$ of $(0,0)$ such that $g(N \times M) \subset V$. Since $G \times G$ has the product topology, $N \times M$ is a neighborhood of $0$ if and only if $N$ and $M$ are neighborhoods of $0$. Hence $x,y \in N$ and $x,y \in M$ and hence $g((x,y)) = x + y \in V$ since $g(N \times M) \subset V$.

$\def\cl{\mathop{\mathrm{cl}}}$ For $x \in G$, let $\mathcal U_x$ denote the set of all neighbourhoods of $x$. Then we have that $x - \mathcal U_0 = \mathcal U_x$ for each $x \in G$. It follows \begin{align*} x \in \cl\{0\} &\iff \forall U \in \mathcal U_x : U \cap \{0\} \ne \emptyset\\ &\iff \forall V \in \mathcal U_0: (x - V) \cap \{0\} \ne \emptyset\\ &\iff \forall V \in \mathcal U_0: 0 \in x - V\\ &\iff \forall V \in \mathcal U_0 : x \in V\\ &\iff x \in \bigcap \mathcal U_0. \end{align*}