I am new to the study of topological vector space and I would like to show the result stated in the title above but I have some doubt, hence the question.
For the first implication, I don’t know how use the fact that the topological vector space (TVS) is separated. Since the mapping $f(x)=\lambda x$, $\lambda\neq 0$ is an homeomorphism it seems natural for me to conclude by noting that $f^{-1}(0_{\mathbb{R}})=\{0\}$ is closed as the inverse image of a closed set by a continuous function.
For the second implication, I have no clue on how to proceed. If you have some hint (rather than full answer) I would be glad to read this please.
Do not hesitate to tell me what goes wrong in my first implication also.. thank you a lot !
Of course $f$ is not real valued so the first implication is false as noticed by a comment !
See the comments for your idea about the left-to-right implication: $x \mapsto \lambda x$ is not a map from the Hilbert space to $\Bbb{R}$.
As for hints: if $0$ is closed, then for any $v \neq 0$, $0$ has an open neighbourhood $U$ such that $v \not\in U$. What can you say about $U + v = \{ u + v : u \in U\}$? If $0$ is not closed, then there is a $v \neq 0$, such that every neighbourhood of $v$ contains $0$.