If $U$ is a subspace of a normed space $X$, must points outside $U$ be a nonzero distance from $U$?

Question

Let $X$ be a normed space over $\mathbb K=\mathbb R$ or $\mathbb C$ and $U\subsetneq X$ be a proper subspace. Suppose further that $X\ni x \notin U$.

1. Must we have $d(x,U)>0$ in general?
2. Must we have $d(x,U)>0$ if $\dim U<\infty$?

Note that $d(x,U):=\inf_{u\in U} \| x-u \|$.

I'm pretty sure the answer to (2) is "yes," since in this case $X\simeq\mathbb K^{\dim U}$, and then we can use topological theorems about $\mathbb K^n$ (as in this related question). Even so, I'd like to see a simpler proof that doesn't rely on the isomorphism with $\mathbb K^n$, since this seems like unnecessary baggage.

Answer 1

$d(x,U)=0$ if and only if $x \in \overline U$. So a dense porper subspace gives a counter-example to the first part. For a specific example take $U=\ell^{0}$ in $X=\ell^{2}$. Since finite dimensional subspaces are closed we have $d(x,U) >0$ when $U$ is finite dimensional.

Answer 2

Hints:

1. take $X$ to be a space of convergent sequences and $U$ to be the subspace of $X$ comprising sequences that are finitely non-zero. This shows that claim (1) is false.

2. choose a basis $\langle u_1, \ldots u_k \rangle$ for $U$ and extend it to the basis $\langle u_1, \ldots u_k, x \rangle$ for $\mathrm{span}(U, x)$. If $d(x, U) = 0$, then $x \in U$. (This is essentially the argument that you don't like because of what you describe as excess baggage, but there is no way of avoiding it.)