A proper subspace of a normed vector space has empty interior. In a vector normed space $E$, prove that all vectorial subspace $F\neq E$ has a interior empty. 
My approach:We consider, the open ball $B\subset F$, with $F$ proper subspace of $E$. If $x\notin F$, then we can increase the radius of the open ball $B$, such that $x\in B$, then the open ball $B$ came out of the subspace $F$, therefore there can be no such open ball. This is my intuition, but I don't know how write this "mathematically". 
 A: Setup and a hint:
Let $x \in F$. Let $y \in E \setminus F$, $\| y \|=1$. Then $\| x - (x+\varepsilon y/2) \|=\varepsilon/2<\varepsilon$. So if $x+\varepsilon y/2 \not \in F$, then $B_\varepsilon(x) \not \subset F$. 
Now show that for every $\varepsilon > 0$, $x+\varepsilon y/2 \not \in F$. My hint: suppose it is in $F$, and conclude that $y \in F$, which is a contradiction. 
So we have shown that $F$ does not contain any open balls around any of its points. Why does this imply it does not contain any open balls at all?
Your approach also works. Suppose $F$ contains some ball. Then it also contains a ball of the same radius centered at zero (subtract the center from each vector in the original ball). Hence it also contains any dilation of this new ball centered at zero (multiply by whatever scalar you like). But any vector can be in some such dilation; thus $F=E$ which is a contradiction.
I like my approach better, because we do similar tricks in loosely related problems. (For instance, we can use a very similar technique to prove that $L^2[0,1]$ is a meager subset of $L^1[0,1]$.)
