Prove that any $T_1$ topological vector space is regular

Question

Proposition

For a $T_1$ space $(X,\mathscr T)$ the following three conditions are equivalent:

1. X is regular;
2. for any $x\in X$ and any $U$ containing it there exist an open neighborhood of $x$ such that $V\subseteq\overline V\subseteq U$;
3. any $x\in X$ has a local base whose element are closed.

Theorem

Suppose $K$ and $C$ are disjoint subsets of a topological vector space X such that K is compact and C is closed. so there exist a symmetric (open) neighborhood $V$ of $0$ such that $$ (K+V)\cap(C+K)=\emptyset $$

Corollary

Let be $X$ a topological vector space. If $\mathscr B(0)$ is a local (open) base centered at $0$ then any its element contains the closure of another one element.

So knowing these results I ask to me if any $T_1$ topological vector space is regular too and so I arranged the following arguments. If $U$ is an open set containing $x$ then $U-x$ is an open set (any raslation is an homeomorphism!) containing $0$ and so there exist an open neighborhood $V$ of $0$ whose closure is contained $U-x$ so that $V+x$ is an open set containing $x$ whose closure is contained in $U$ and thus the statement follows immediately by the proposition mentioned above. So is the argument true and are my arguments correct? Could someone help me, please?

Answer 1

Your argument is correct. You could also apply the theorem directly: if $x\in U\subseteq X$, where $U$ is open, then $\{x\}$ is compact, and $C=X\setminus U$ is closed and disjoint from $\{x\}$, so there is an open nbhd $V$ of $0$ such that $(x+V)\cap(C+V)=\varnothing$. But then $C+V$ is an open nbhd of $C$, so $F=X\setminus(C+V)$ is a closed subset of $U$, and $x+V$ is an open nbhd of $x$ such that $\operatorname{cl}(x+V)\subseteq F\subseteq U$.