 Solution

If a linear subspace $V$ of a normed space contains an open ball $B(v,r)$, then $v\in V$ and subtraction gives $B(0,r)\subseteq V$. For any $x\in X$, choose a positive scalar $t$ with $\Vert tx\Vert <r$. Then $tx\in V$ and hence $x\in V$. Thus $V=X$. Every proper linear subspace has empty interior.