Let $X$ be a normed space and let $V$ be a nonzero subspace of $X$ which is not dense in $X$. I want to prove that for every $\epsilon>0$ there exists a unit vector $x\in X$ such that $0<d=\inf_{v\in V}\|x-v\|< \epsilon$.
Here is the sketch of my proof (Edit: please ignore this part: under the additional hypothesis that $X$ is a topological vector space.)
Define $f\colon (0, \epsilon)\times V$ by $f(s, v)=\|sp+v\|$ where $p$ is a fixed unit vector in $X\setminus \overline{V}$. Of course, $f$ is continuous with a connected domain. Moreover, the range of $f$ is $(0, \infty)$. Thus $f$ takes the output $1$ for some combination of $s_0$ and $v_0$. Clearly $x=f(s_0, v_0)$ has the desired properties.
Although the solution looks fine, I think that there should be easier/different solutions as well and I would like to see if my argument can be simplified.
Edit: (please ignore this part) I'm also not quite sure if the extra assumption of "topological vector space" for $X$ can be dropped.
Any clarification is appreciated.