Let $X$ a topological vector space. If $A \subset X$ then $\bar{A} = \bigcap (A + V)$, where $V$ runs through all the neighborhoods of $0$.
The proof seems ok until the very last bit.
Basically if $x \in \bar{A}$ then $(x + V) \cap A \neq \emptyset$, for any $V$ neighborhood of $0$, this happens if and only if $x \in A - V$ for every such $V$. After I don't get the meaning of the following
Since $-V$ is a neighborhood of $0$ if and only if $V$ is one, the proof is complete.
What's the meaning of "$V$ is one"?
Here "if $V$ is one" means "if $V$ is also a neighborhood of $0$". So $-V$ is a neighborhood of $0$ if and only if $V$ is a neighborhood of $0$ as well (since $0=-0$).