Finite dimensional topological vector space non complete

Let $V$ be a TVS and denote with $N$ the closure of $\{0\}$, this is again a vector sub-space as it is the closure of a vector sub-space. Further, every open neighbourhood of $0$ contains all of $N$ (and so does every open neighbourhood of any point in $N$). I'll carry this last comment out in more detail, suppose $x,y\in N$, if $U$ is an open neighbourhood of $x$ not containing $y$ you have that $U-y$ is an open neighbourhood of $x-y$ not containing $0$. But $x-y\in N$ and $N$ is the closure of $\{0\}$ and every open neighbourhood of points in $N$ must intersect $\{0\}$, contradiction.

Now $V/N$ is a quotient of a topological group by a closed sub-group, hence it is Hausdorff by general facts. In other words $V/N$ is a Hausdorff TVS, since it is also finite dimensional it is complete.

So let $x_\alpha$ be a Cauchy net in $V$, meaning that for any neighbourhood $U$ of $0$ you have a $\gamma$ so that $x_\alpha - x_\beta\in U$ for all $\alpha, \beta ≥\gamma$. By continuity of the projection map you have that $[x_\alpha]$ has the same property in $V/N$, hence admits a limit $[x]$ by completeness. We will now see that any lift $x$ of $[x]$ is a limit of $x_\alpha$.

As we have seen $[x_\alpha - x]$ eventually lies in every neighbourhood of $[0]$, meaning that $x_\alpha-x+N$ eventually lies in $U+N$ for every neighbourhood $U$ of $0$. But by what we have seen in the first paragraph $U\supseteq N$ hence $x_\alpha - x$ eventually lies in $U$. This shows $x_\alpha\to x$.

Denote by $\mathscr{F}_E(y)$ be the filter of neighborhoods of a point $y$ in the TVS $E$.

Let $V$ be a finite dimensional TVS and denote with $N$ the closure of $\{0\}$, this is again a vector sub-space as it is the closure of a vector sub-space. Further, every open neighbourhood of $0$ contains all of $N$.

Now $V/N$ is a quotient of a topological group by a closed sub-group, hence it is Hausdorff space. In other words $V/N$ is a Hausdorff TVS, since it is also finite dimensional it is complete.

So let $\mathscr{F}$ be a Cauchy filter in $V$. By continuity of the projection map we have that $\phi(\mathscr{F})$ is a Cauchy filter in $V/N$, hence admits a limit $\overline{x}$ by completeness. We will now see that any lift $x$ of $\overline{x}$ is a limit of $\mathscr{F}$ in $V$.

Let $Z \in \mathscr{F}_{V/N}(\overline{x})$, since $\mathscr{F}_{V/N}(\overline{x}) \subset \phi(\mathscr{F})$, there exists $F \in \mathscr{F}$ such that $\phi(F) \subset Z$. Therefore, $F \subset \phi^{-1}(\phi(F)) \subset \phi^{-1}(Z)$, which implies that $\phi^{-1}(Z) \in \mathscr{F}$.

We claim that $\mathscr{F}_V(x) \subset \mathscr{F}$. Let $U+x \in \mathscr{F}_V(x)$ and consider $W \in \mathscr{F}_V(0)$ such that $W+W \subset U$. Define $Z=\phi(W+x)=\phi(W)+\overline{x} \in \mathscr{F}_{V/N}(\overline{x})$. As we have seen $\phi^{-1}(Z) \in \mathscr{F}$, that is, $W+x+N=\phi^{-1}(\phi(W+x)) \in \mathscr{F}$. Observing that $W+x+N \subset W+x+W \subset U+x$, we obtain $U+x \in \mathscr{F}$ as desired.