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.

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 $f(B)=\mathbb R$.
