Vector Spaces and Open Balls 
Let $X$ be a vector space, equipped with a norm $\Vert\cdot\Vert$. Let $V$ be a subspace of $X$.
Suppose that $V$ contains some open ball (open ball of ($X$, $\Vert\cdot\Vert$)) centered at $0$. Show that $V = X$.

I think I have an intuition for how to prove this. A ball of ($X, d$) centered at $0$ contains all possible directions in $X$. If a vector space contains it, then it contains all "multiples" of all directions, which means it has to be the whole space. I'm can't really seem to put this into a formal proof, though.
I could take some $x \in X$. If I find some scalar multiple of $x$ in the open ball, and therefore in $V$, then it follows that $x \in V$, which means $X$ = $V$. But how do I find this scalar multiple? Do I take some arbitrary $\alpha$? Since the open ball contains all possible directions, I think any arbitrary scalar would do, but I'm not quite sure.
I could then generalize the first part with the fact that $B_{\Vert\cdot\Vert} (x, r) = x + B_{\Vert\cdot\Vert} (0, r)$ to show the second part. But it's the first part I'm having trouble formalizing into a proof. Thank you for your help.
 A: For the first one, suppose that $B(0,r)\subset V$, for every $x\in X, x\neq 0$, $u={r\over 2}{x\over{\|x\|}}\in V$, this implies that $x={{2\|x\|}\over r}u\in V$.
For the second, suppose that $B(x,r)\subset V$, let $u\in B(0,r)$, $x+u\in B(x,r)\subset V$, we deduce that $x+u\in V$, since $V$ is a vector space, $u=(x+u)-x\in V$, we deduce that $B(0,r)\subset V$ we can apply the first part.
A: Hint: think about lines through the centre of the open ball (in the first case) and through the origin as well (in the second case): this reduces the first part to the 1-dimensional case and the second part to the 2-dimensional case.
A: For your scalar multiple, you're correct that all you have to do is choose an arbitrary $c \in (0,r)$ and multiply the vector $x$ by $\frac{c}{\lVert x \rVert}$. One easy option would simply be $c = \frac{r}{2}$, and the resulting rescaled vector would have to be in $B(0,r)$. I'm not sure what the second part you're referring to is, but if it is proving the statement for an open ball centered at some arbitrary point in $V$, you've got the right idea.
