Unbounded linear functional maps every open ball to $\mathbb{R}$? I can't get my head wrapped around this:
Let $X$ be a normed linear space. Let $f:X\rightarrow\mathbb{R}$ be a linear functional on $X$. Prove that $f$ is unbounded if and only if $\forall y\in X$ and $\forall \delta>0$ we have $\{f(x)\,:\,|x-y|<\delta\}=\mathbb{R}$.
I have already proved:
1.) $f$ is continuous if and only if $f$ is bounded
2.) $f$ is bounded if and only if $f^{-1}(0)$ is closed
3.) either $f^{-1}(0)$ is closed ($f$ bounded) or $f^{-1}(0)$ is dense in $X$ ($f$ unbounded)
I know that all these ideas play off one another in some fashion, but cannot seem to tease out a solution to the above statement. Any help or direction would be appreciated.
 A: Jonathan's suggestion is spot on, but let me give you a more explicit argument. Consider first the unit ball $X_1$ of $X$. As $f$ is unbounded, there exists a sequence $\{x_n\}\subset X_1$ with $|f(x_n)|>n$. By replacing $x_n$ with $-x_n$ if necessary, we can get $f(x_n)>n$. 
Given $t\in[0,1]$, $tx_n\in X_1$, and $f(tx_n)=tf(x_n)$. This shows that that $f(X_1)$ contains the whole segment $[0,f(x_n)]$, and in particular $[0,n]$. As we can do this for all $n$, we get that $f(X_1)\supset[0,\infty)$. Finally, using that $X_1=-X_1$, we get $f(X_1)=\mathbb R$. 
Any other ball $B$ in $X$ is of the form $x+\delta X_1$ for some $\delta>0$. Given such $\delta$, for any fixed $t\in\mathbb R$ by the previous paragraph there exists $y\in X_1$ with $f(y)=(t-f(x))/\delta$. Then $f(x+\delta y)=t$. So $f(B)=\mathbb R$.
A: To go along with Jonathan's suggestion and Martin's technical answer, I'd like to give some intuition.
Since $f$ is unbounded, there is some bounded set, $\Omega$, which is taken to an unbounded set in $\mathbb{R}$. Since $\Omega$ is bounded, it can be enclosed in a large ball, $B(0,R)$. This ball maps onto all of $\mathbb{R}$. Therefore any scaling of it maps onto $\mathbb{R}$, so any radius ball about the origin must map onto all of $\mathbb{R}$. 
Now any ball in $X$ is a translate of a ball about the origin; as Martin suggests in his answer, it's just a change of coordinates to see that this must also map onto all of $\mathbb{R}^n$. (Morally speaking, for a linear map, what happens near zero must happen near every point, although unbounded maps are not differentiable.)

Here are some details.
Suppose $\Omega$ maps to an unbounded set in $\mathbb{R}$ and  The image of $B(0,R)$ must be all of $\mathbb{R}$, because for any $y\in\mathbb{R}$ there is some $x\in\Omega$ such that $f(x) > y$; put $y/f(x) = r$, so that $f(rx) = y$ and since $r<1$ and $x\in B(0,R)$, $rx\in B(0,R)$.
Now for any $\epsilon > 0$, $B(0,\epsilon R)$ maps onto all of $\mathbb{R}$. To see this, observe that for any $y\in\mathbb{R}$, there is some $x\in B(0,R)$ which maps to $y/\epsilon$. Therefore $\epsilon x\mapsto y$, and $\epsilon x\in B(0,\epsilon R)$. 
Now apply the coordinate-change argument in Martin's answer.
