# Closure of intersection with vector subspace

I am confused with the footnote on page 198 of http://www.ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf

Essentially:

Let $X$ be a topological vector space and $Y$ a finite-dimensional vector subspace. Let $V$ be an open (balanced) subset of $X$ containing $0$. Then $V \cap \overline{Y} \subseteq \overline{V \cap Y}$.

In general, it is not true that $\overline{V} \cap \overline{Y} \subseteq \overline{V \cap Y}$ for arbitrary subsets $V$ and $Y$ (any topological space that has this property must be discrete) so the obvious approach doesn't work

Suppose that $x\in V\cap\operatorname{cl}Y$, and let $U$ be an open nbhd of $x$. $V$ is an open nbhd of $x$, so without loss of generality we may assume that $U\subseteq V$. Then $U\cap Y\ne\varnothing$, since $x\in\operatorname{cl}Y$, so
$$U\cap(V\cap Y)=U\cap Y\ne\varnothing\;,$$
and $x\in\operatorname{cl}(V\cap Y)$.