A second proof of the fact that a linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed. I am working through Folland, not for homework, and have come across this problem (Ch 5 #17):

A linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed.

Note the "only if" direction is trivial. Folland suggests that we use the following fact in our proof: for all $\epsilon>0$ there exists an $x\in\mathcal{X}$ such that $\|x\|=1$ and $\|x+\mathcal{M}\|\ge 1-\epsilon$, where $\mathcal{M}$ is a proper closed subspace of $\mathcal{X}$.
I can prove both the hint, and the exercise, but not the latter using the former! My proof is to suppose (toward a contradiction), that there existed some sequence $\|x_n\|=1$ such that $f(x_n)>n$ for each $n\in\mathbb{N}$, and then consider the sequence $\{x-f(x)\frac{x_n}{f(x_n)}\} $ and observe that this is in the kernel of $f$ for each $n$ and approaches $x$ in the limit, from which the statement should follow since $x$ is arbitrary. 
Can anyone tell me how to use the hint? I'm guessing the closed subspace we consider is the kernel of $f$, but it's not clear how to go from there.
Any help would be appreciated.
 A: The hint gives you an $x\in X$, with $\|x\|=1$, and such that $x\not\in\ker f$ (because this is equivalent to $\|x+\ker f\|\ne0$). Now, since $f(x)\ne0$ we can write, for any $y\in X$,
$$
y=\frac{f(y)}{f(x)}\,x+\left(y-\frac{f(y)}{f(x)}\,x\right),
$$
where the term in brackets belongs to $\ker f$. So $X=\mathbb{C}x\oplus\ker f$. If we write $k=|f(x)|$ we get, for any $\lambda x+m\in\mathbb{C}x\oplus\ker f$,
$$
|f(\lambda x+m)|=|f(\lambda x)|=|\lambda|\,|f(x)|=k\,|\lambda|.
$$
From the hint, we also have
$$
\|\lambda x+m\|=|\lambda|\,\|x+\lambda^{-1}\,m\|\geq|\lambda|\,(1-\varepsilon).
$$
Joining the two estimates we get
$$
|f(\lambda x+m)|=k\,|\lambda|\leq\,\frac{k}{1-\varepsilon}\,\|\lambda x + m\|.
$$
So $f$ is bounded.
A: The linear map $f$ factor through the quotient  $F: X/\ker(f)\to \mathbb R$ and it is continuous since is a linear map between finite dimensional subspaces. On the other hand, the quotient map $\pi:X/\ker(f)\to \mathbb R$ is continuous as $\ker(f)$ is closed, hence $f=F\circ \pi$ so it is a composition of continuous functions. 
