finite dimensional proper subspaces of a normed space are nowhere dense

Proposition - Let $V$ be a normed space. If $S \subseteq V$ is a finite dimensional proper subspace, then $S$ is nowhere dense.

Proof :

It is known that for any topological vector space (in particular, normed spaces) every proper subspace has empty interior.

From the parent entry we also know that finite dimensional subspaces of $V$ are closed.

Then, $\operatorname{int}(\overline{S}) = \operatorname{int}(S) = \emptyset$ , which shows that $S$ is nowhere dense. $\square$