One dimensional vector space and not Hausdorff I read that all vector spaces that do not have the Hausdorff property and are one-dimensional need to have the trivial topology. I am not quite sure how to approach this problem, but I would like to get an idea, how we can show this?-Somehow I assume that this problem should not be too hard, so I would be glad about any possible hint.
Thus, either full solutions or ideas are highly appreciated.

Suppose $E$ is a topological vector space over a field $K$ who does not have trivial topology, i.e there is an open set $O$ different from $E$ and $\emptyset$. Choose $x\in O$ and $y\notin O$. Consider, for all $k\in K^{*}$, the continuous function $f_{k}:v\mapsto x+k^{-1}(v-x)$. Then $O_{k}=f_{k}^{-1}(O)$ is open, contains $x$, and does not contain $x+k(y-x)$ since $f_{k}(x+k(y-x))=y$.
If $E$ is one-dimensional, every $z\in E\setminus\{x\}$ can be written $x+k_{0}(y-x)$ for some $k_{0}$, and then $O_{k_{0}}$ separates $x$ from $z$. This shows $x$ can be separated from every other point in $E$. To show that for general $x'\in E$ do the reasoning above again with $O'= O+(x'-x)$, which is open again by continuity of addition.
